Main.java:7: error: incompatible types: String cannot be converted to int
        int hours = scanner.nextLine();
                                    ^
1 error

The good news: this is a compile error. Java found the problem before running anything, and it tells you the exact file, line and spot. Here's what it means, the usual causes, how to fix each one, and the related errors you're likely to meet next.

What the error means

• Main.java:7: the file and the line number.
• incompatible types: you put a value of one type into a place meant for another type.
• String cannot be converted to int: the value is a String (text), and the place, here the variable hours, is an int (a whole number). Java won't turn one into the other on its own.

The ^ mark shows where Java noticed it: at the end of scanner.nextLine(), the expression that produced the text.

Java checks types very strictly. A variable declared as int can only ever hold a whole number, so Java stops you as soon as you try to put text in it.

The same error with java.lang.String

Depending on your editor, the same message can look a little different, for example:

java: incompatible types: java.lang.String cannot be converted to int

java.lang.String is simply the full name of String (the class lives in the package java.lang), and the java: at the start only says that the message comes from the Java compiler. It's the same error, with the same causes and the same fixes.

Cause 1: reading a number with nextLine()

This is the most common one. You ask a question with a Scanner and want the answer as a number:

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        System.out.print("How many hours a day are you on your phone? ");
        int hours = scanner.nextLine();
        System.out.println("That is " + hours * 365 + " hours a year.");
    }
}

scanner.nextLine() always gives back the whole line the user typed as a String, even if they typed only digits like 3. The text "3" isn't the number 3, so it can't go into an int. The same goes for scanner.next(), which reads one word: it also gives back a String.

The fix: Integer.parseInt

Read the line as text, then convert it with Integer.parseInt:

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        System.out.print("How many hours a day are you on your phone? ");
        int hours = Integer.parseInt(scanner.nextLine());
        System.out.println("That is " + hours * 365 + " hours a year.");
    }
}

How many hours a day are you on your phone? 3
That is 1095 hours a year.

Read it from the inside out: nextLine() gives the text "3", Integer.parseInt turns it into the number 3, and that number goes into hours.

What about nextInt()?

Scanner also has nextInt(), which reads a number directly, so it seems like the simpler fix. It works, but it has a trap that catches almost everyone. Ask for a number with nextInt() and then for a name with nextLine():

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        System.out.print("Hours a day? ");
        int hours = scanner.nextInt();
        System.out.print("Your name? ");
        String name = scanner.nextLine();
        System.out.println("Name: [" + name + "], hours: " + hours);
    }
}

Type 3, press Enter, and the program doesn't wait for your name at all:

Hours a day? 3
Your name? Name: [], hours: 3

nextInt() reads the 3 but leaves the Enter you pressed behind. The next nextLine() finds that leftover line break straight away and returns an empty line. That's why reading every answer with nextLine() and converting with Integer.parseInt is the safer habit. (If you do use nextInt(), call scanner.nextLine(); once right after it to throw away the rest of the line.)

Cause 2: a value in quotes

The same error appears when you write a value in double quotes by mistake. Anything in double quotes is a String:

int age = "thirty";

Main.java:3: error: incompatible types: String cannot be converted to int
        int age = "thirty";
                  ^
1 error

Here the ^ points at the text itself. The fix depends on what you meant: a number goes without quotes, int age = 30;, and if you really want text, the variable must be a String: String age = "thirty";.

Cause 3: a method that returns a String

A method's return type decides what its result is, wherever you use it. If a method is declared to return a String, you can't put its result into an int:

public class Main {
    static String askName() {
        return "Lena";
    }

    public static void main(String[] args) {
        int count = askName();
    }
}

Main.java:7: error: incompatible types: String cannot be converted to int
        int count = askName();
                           ^
1 error

Look at the method's first line, static String askName(): the String before the name is what it gives back. Either store the result in a String, or, if the method should really give a number, change its return type and what it returns.

What can happen next: NumberFormatException

Once the program compiles, try typing 2.5 as the answer:

How many hours a day are you on your phone? 2.5
Exception in thread "main" java.lang.NumberFormatException: For input string: "2.5"
	at java.base/java.lang.NumberFormatException.forInputString(NumberFormatException.java:67)
	at java.base/java.lang.Integer.parseInt(Integer.java:668)
	at java.base/java.lang.Integer.parseInt(Integer.java:786)
	at Main.main(Main.java:7)

This is a different kind of problem. The code compiled fine, but while running, Integer.parseInt got text that isn't a whole number. That's called an exception, and the last at line shows where it came from in your code: Main.java:7.

You have two options:

• If decimals are allowed, use a double and Double.parseDouble instead.
• If you need a whole number, catch the exception and ask nicely:

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        System.out.print("How many hours a day are you on your phone? ");
        String text = scanner.nextLine();
        try {
            int hours = Integer.parseInt(text);
            System.out.println("That is " + hours * 365 + " hours a year.");
        } catch (NumberFormatException e) {
            System.out.println("Please type a whole number, like 3.");
        }
    }
}

How many hours a day are you on your phone? 2.5
Please type a whole number, like 3.

Related errors

The same "incompatible types" message comes in a few other flavours. Each one names the two types, so read them the same way: the value's type first, then the type of the place it's going.

int cannot be converted to String

The opposite direction:

int hours = 3;
String label = hours;

Main.java:4: error: incompatible types: int cannot be converted to String
        String label = hours;
                       ^
1 error

Turn the number into text with String.valueOf(hours), or join it to some text with +, which makes the whole thing a String:

String label = String.valueOf(hours);   // "3"
String sentence = "Hours: " + hours;    // "Hours: 3"

String cannot be converted to double

The same problem with a decimal number, for example double price = scanner.nextLine();. The fix is the decimal version of parseInt:

double price = Double.parseDouble(scanner.nextLine());

String cannot be converted to char

char grade = "A";

Main.java:3: error: incompatible types: String cannot be converted to char
        char grade = "A";
                     ^
1 error

In Java, double quotes make a String, and single quotes make a single character, a char. Write char grade = 'A';. If you have a String and want its first character, use text.charAt(0).

possible lossy conversion from double to int

int total = 7;
int average = total / 2.0;

Main.java:4: error: incompatible types: possible lossy conversion from double to int
        int average = total / 2.0;
                            ^
1 error

7 / 2.0 is 3.5, a double, and putting it into an int would lose the .5. Java won't do that silently. Either keep the decimals with double average = total / 2.0;, or say that you really want to cut them off with a cast: int average = (int) (total / 2.0);, which gives 3.

Compile errors and exceptions

It helps to keep the two apart:

• A compile error like incompatible types stops javac before anything runs. It always has error: and a line number in your file.
• An exception like NumberFormatException happens while the program runs, often because of the input. It starts with Exception in thread "main".

Both point you to a line in your code. Start there. For the exception you'll probably meet soon after this one, see Java NullPointerException. And if you also write Python, it has its own version of this mix-up: Python TypeError: unsupported operand type(s) for -: 'int' and 'str'.

A quick checklist

1. Read the two types in the message: the first is the value you have, the second is what the variable (or parameter) expects.
2. Find where the value comes from: nextLine() and next() always give a String, so does anything in double quotes, and so does a method declared to return String.
3. Convert on purpose: Integer.parseInt or Double.parseDouble for text to numbers, String.valueOf for numbers to text, single quotes for a char.
4. Prefer nextLine() plus parseInt over nextInt(), or clear the line after nextInt().
5. Once it compiles, test with input you didn't expect, like 2.5 or a word, and handle the NumberFormatException.

The same way of reading errors works in other languages too: How to Read an Error Message: A 3-Step Method for Python, Java, JavaScript and SQL.

This error is part of our free Java trial, "Your Year on Screen": you read an answer with scanner.nextLine() into an int, and the compiler stops you. The AI tutor doesn't hand you Integer.parseInt. It first asks you to read the error and say which two types it talks about, and gives the next hint only if you need it. One free 20-minute session, nothing to install: craftstackacademy.com/trial.

This article was first published on the Craftstack Academy blog.

Traceback (most recent call last):
  File "main.py", line 3, in <module>
    age = 2026 - year
          ~~~~~^~~~~~
TypeError: unsupported operand type(s) for -: 'int' and 'str'

This is one of the most common errors in a first Python program. The fix is one word, but it's worth understanding why it happens, because the same idea comes back again and again.

Here is the program:

# main.py
name = input("Your name? ")
year = input("Born in? ")
age = 2026 - year
print("Hi", name, "- you are", age)

What the error means

Read the last line of the error from the left:

• TypeError: the problem is the type of a value, not its spelling.
• unsupported operand type(s) for -: the minus sign got values it can't work with.
• 'int' and 'str': the value on the left is an int (a whole number, 2026) and the value on the right is a str (a string, which means text).

Python can subtract a number from a number. It can't subtract text from a number, and it doesn't guess that the text "2002" was meant as a number.

The lines above the last one tell you where it happened: line 3 of main.py. The ^ mark points at the minus sign. (Python 3.11 and newer draw that mark; older versions only show the line.)

Why year is text

input() always gives back a string, even when you type only digits. You can check it yourself with type():

year = input("Born in? ")
print(type(year))

Born in? 2002
<class 'str'>

So year holds the text "2002", not the number 2002. To Python those are as different as the word "five" and the number 5.

The fix: turn the text into a number

Wrap the input() in int():

# main.py
name = input("Your name? ")
year = int(input("Born in? "))
age = 2026 - year
print("Hi", name, "- you are", age)

Your name? Lena
Born in? 2002
Hi Lena - you are 24

Read the new line from the inside out: input() asks the question and gives back text, then int() turns that text into a whole number. Now both sides of the minus sign are numbers.

For numbers with a decimal point, like a height of 1.75, use float() instead of int().

The same error with other operators, and the other order

The two types in the message are in the same order as in your code: the value left of the operator first, then the value right of it. So if the text is on the left, the message turns round:

year = "2002"
next_birthday = year - 1

TypeError: unsupported operand type(s) for -: 'str' and 'int'

The symbol after for tells you which operator failed. With text on the left, you'll see the same message for division and floor division, and a slightly longer one for powers:

TypeError: unsupported operand type(s) for /: 'str' and 'int'
TypeError: unsupported operand type(s) for //: 'str' and 'int'
TypeError: unsupported operand type(s) for ** or pow(): 'str' and 'int'

And with a number on the left of a +, like 2026 + year, it's for +: 'int' and 'str'. You get the same message from sum() on a list of strings, such as sum(["72", "85"]), because sum adds the items up one by one, starting from the number 0.

The fix is the same every time: convert the text to a number before you calculate, with int() or float().

Three operators behave differently, so watch out for them:

• year + 1 with text on the left gives can only concatenate str (not "int") to str (see below).
• year % 2 gives the puzzling TypeError: not all arguments converted during string formatting, because % after a string means "fill in this text", not "remainder".
• year * 2 gives no error at all: it prints 20022002. Multiplying text by a whole number repeats the text. (Two strings, like "5" * "3", do fail: can't multiply sequence by non-int of type 'str'.)

If the message says 'NoneType' instead of 'str', the value isn't text but nothing: often a function that prints its result instead of returning it.

Two errors you may meet next

can only concatenate str (not "int") to str

This is the same problem the other way round. + between two strings joins them, but here one side is a number:

print("Born in " + 2002)

TypeError: can only concatenate str (not "int") to str

Either let print() put the pieces together with commas, print("Born in", 2002), or turn the number into text first with str(2002).

ValueError: invalid literal for int() with base 10

int() only works when the text really is a whole number. If someone types a word, the conversion fails:

Your name? Lena
Born in? nineteen
Traceback (most recent call last):
  File "main.py", line 2, in <module>
    year = int(input("Born in? "))
ValueError: invalid literal for int() with base 10: 'nineteen'

This one isn't a mistake in your code: it's input your program didn't expect. Handling it with try and except is the topic of Error Handling in Python Course 3.

Another error from a first program that's worth knowing: Python IndentationError: expected an indented block after 'if', 'for' or a function. And if you're also learning Java, it has its own version of this mix-up: Java: incompatible types: String cannot be converted to int.

How to read any TypeError

1. Find the line number in the traceback, and the ^ mark under the line.
2. Look at the operator or function it points to: -, +, len() and so on.
3. Read the two types the message names. If you're not sure which value is which, print type() of each.
4. Convert one side so the types match: int(), float() or str().

The same way of reading errors works in other languages too: How to Read an Error Message: A 3-Step Method for Python, Java, JavaScript and SQL.

This exact error is part of our free Python trial, "Your Life in Numbers". When it happens, the AI tutor doesn't tell you to add int(). It first asks you to read the last line and say which two types it mentions, and only gives the next hint if you need it, so the fix is one you found yourself. One free 20-minute session, nothing to install: craftstackacademy.com/trial. It leads into our Python course.

This article was first published on the Craftstack Academy blog.

SELECT name FROM visitors WHERE room = Vault;

and SQLite answers:

Error: no such column: Vault

But Vault isn't supposed to be a column. It's the value you're looking for. That's exactly the problem, and the fix is two small marks.

Try it yourself

Here's a tiny table to run in any SQLite tool, such as DB Browser for SQLite or the SQL editor in a Craftstack session:

CREATE TABLE visitors (id INTEGER PRIMARY KEY, name TEXT, room TEXT, entered TEXT);
INSERT INTO visitors (name, room, entered) VALUES
  ('Ava Brooks', 'Lobby', '21:40'),
  ('Ben Carter', 'Vault', '22:15'),
  ('Cleo Diaz', 'Vault', '22:50'),
  ('Dan Evans', 'Cafe', '23:05');

Why SQLite looks for a column

Apart from SQL's own keywords like SELECT and WHERE, a word without quotes is a name: the name of a table, a column or a function. So in WHERE room = Vault, SQLite reads Vault the same way it reads room, as a column, and compares the two columns. There's no column called Vault, so it stops.

Text values, the actual words stored in the table, go in single quotes:

SELECT name FROM visitors WHERE room = 'Vault';

Ben Carter
Cleo Diaz

Numbers don't need quotes (WHERE id = 2), and neither do column names. Text always does.

What about double quotes? SQLite often accepts "Vault" too, but only as a fallback: double quotes are meant for names, and other databases such as PostgreSQL read "Vault" as a column. Use single quotes for text and your queries will work everywhere.

The tool you use may word the start of the message differently. Python's sqlite3 module, for example, raises sqlite3.OperationalError: no such column: Vault. The important part is the same: no such column, followed by the word SQLite couldn't find.

Four other causes

A typo in a column name

SELECT nmae FROM visitors;

Error: no such column: nmae

The word at the end of the message is your clue: it's exactly what you typed. Compare it with the real column names letter by letter.

A column that's in another table

SELECT name, coat FROM visitors;

Error: no such column: coat

Maybe coat exists, but in a different table. A query can only use columns from the tables in its FROM part. To use columns from two tables, you join them with JOIN, which is a topic of its own.

A table alias that wasn't declared

Short names for tables, like v for visitors, only exist if you declare them in the FROM part:

SELECT v.name FROM visitors WHERE room = 'Vault';

Error: no such column: v.name

Add the alias after the table name, and use it consistently:

SELECT v.name FROM visitors AS v WHERE v.room = 'Vault';

A column added after the table was created

This one catches people who build a small app. The code starts with CREATE TABLE IF NOT EXISTS, and later you add a column to it:

CREATE TABLE IF NOT EXISTS visitors (id INTEGER PRIMARY KEY, name TEXT, room TEXT, entered TEXT);

If the database file already has a visitors table from before, without entered, this line does nothing: the table exists, so it isn't created again. The next SELECT entered FROM visitors fails with no such column: entered, even though the column is right there in your code.

Add the column to the existing table:

ALTER TABLE visitors ADD COLUMN entered TEXT;

Or, while you're still practising and the data doesn't matter, delete the database file (or DROP TABLE visitors;) and let the code create it again.

Errors that look alike

A few other messages come from the same kind of mix-up between names and values:

• no such table: visitor: the same problem one level up, a misspelled table name (here visitor instead of visitors), or a table that was never created in this database file.
• table visitors has no column named nmae: what SQLite says when the misspelled column is in an INSERT, like INSERT INTO visitors (nmae, room) VALUES ('Eve Fox', 'Lobby');.
• ambiguous column name: name: the opposite problem. After a JOIN, two tables both have a column called name, and SQLite doesn't know which one you mean. Put the table name in front: SELECT visitors.name, rooms.floor FROM visitors JOIN rooms ON visitors.room = rooms.name;.

Once you start counting and grouping rows, a different family of messages appears: SQL GROUP BY errors: "misuse of aggregate" and "must appear in the GROUP BY clause".

The same error in PostgreSQL

If you move to PostgreSQL, the missing quotes give ERROR: column "vault" does not exist. Notice the lower-case vault: PostgreSQL turns names without quotes into lower case, so the message doesn't even show the word exactly as you typed it. And "Vault" in double quotes, which SQLite quietly accepts as text, is a column name in PostgreSQL: ERROR: column "Vault" does not exist. Single quotes work in both.

How to see which columns a table has

In SQLite, PRAGMA table_info lists the columns of a table:

PRAGMA table_info(visitors);

Most tools also show the columns in a sidebar. Check the list before you go hunting for anything more complicated.

A quick checklist

1. Read the word at the end of the message: that's what SQLite tried to use as a column.
2. Is it a value, like a room, a name or a color? Put it in single quotes.
3. Is it meant to be a column? Check the spelling against the table's columns with PRAGMA table_info.
4. Is the column in another table? Add that table with a JOIN.
5. Does it start with a short name like v.? Declare that alias in the FROM part.
6. Did you add the column to your code later? Add it to the existing table with ALTER TABLE ... ADD COLUMN.

Once it works, you can keep adding conditions with AND:

SELECT name FROM visitors WHERE room = 'Vault' AND entered >= '22:30';

Cleo Diaz

The same "read the last word" habit works for errors in other languages too: How to Read an Error Message: A 3-Step Method for Python, Java, JavaScript and SQL.

New to SQL and wondering how it compares with other languages? See Which programming language should you learn first?

Our free SQL trial, "The Museum Mystery", is built around a query just like this one: a diamond is missing from the Vault, and you narrow the visitor log down, clue by clue, to the thief. If you meet "no such column" on the way, the AI tutor lets you find the missing quotes yourself. One free 20-minute session, nothing to install: craftstackacademy.com/trial. It leads into our Databases & SQL course.

This article was first published on the Craftstack Academy blog.

Uncaught TypeError: Cannot read properties of null (reading 'addEventListener')

The message sounds technical, but it says something simple: JavaScript didn't find your element. Here's how to read it, the mistakes that cause it almost every time, and a few related errors that look alike.

What the error means

Read it from the end:

• (reading 'addEventListener'): your code tried to use addEventListener on something.
• of null: that something was null, which means "nothing".
• Cannot read properties: you can't call anything on nothing.

So the variable you called addEventListener on is null. Where did it come from? Almost always from a line like this, just above:

// app.js
const button = document.querySelector("#color-btn");

document.querySelector gives back the first element that matches the selector, or null if nothing on the page matches. Then the next line fails:

// app.js
button.addEventListener("click", () => {
  document.body.style.background = "lightyellow";
});

The error also stops your script at that line, so nothing written after it runs either.

Cause 1: a missing # in the selector

querySelector uses CSS selectors. An id needs a # in front, and a class needs a .:

<!-- index.html -->
<button id="color-btn">Change color</button>
<script>
  const button = document.querySelector("color-btn");
  button.addEventListener("click", () => {
    document.body.style.background = "lightyellow";
  });
</script>

Without the #, "color-btn" means "an element whose tag name is color-btn", like <color-btn>. There's no such tag on the page, so you get null.

The fix is one character:

const button = document.querySelector("#color-btn");

Or use getElementById, which takes the id without the #:

const button = document.getElementById("color-btn");

Check the spelling too: color-btn in the HTML and colour-btn in the JavaScript are different ids, and the result is the same error.

Cause 2: the script runs before the element exists

This version has a perfect selector and still fails:

<!-- index.html -->
<!DOCTYPE html>
<html>
  <head>
    <script src="app.js"></script>
  </head>
  <body>
    <button id="color-btn">Change color</button>
  </body>
</html>

The browser reads the page from top to bottom. It reaches the <script> in the <head> and runs app.js right away, before it has read the <body>. At that moment the button doesn't exist yet, so querySelector returns null.

You have three ways to fix it. Any one is enough:

1. Move the <script> to the end of the <body>, just before </body>.
2. Keep it in the <head> but add defer: <script src="app.js" defer></script>. The browser then runs it after the page has been read.
3. Wrap your code so it waits for the page:

// app.js
document.addEventListener("DOMContentLoaded", () => {
  const button = document.querySelector("#color-btn");
  button.addEventListener("click", () => {
    document.body.style.background = "lightyellow";
  });
});

Cause 3: an id that's really a class

The selector has to use the same kind of name as the HTML. Here the button has a class, but the selector looks for an id:

<!-- index.html -->
<button class="color-btn">Change color</button>
<script>
  const button = document.querySelector("#color-btn");
  button.addEventListener("click", () => {
    document.body.style.background = "lightyellow";
  });
</script>

#color-btn only matches id="color-btn", so querySelector returns null and you get the same error. Match the selector to the HTML: ".color-btn" for a class, "#color-btn" for an id. And remember that getElementById only finds ids: it never finds an element by its class.

Cause 4: one script, several pages

Your site has a home page with the button and an About page without it, and both load the same app.js. On the home page everything works. On the About page, the console shows the error, and anything after that line in app.js doesn't run there either.

The script isn't wrong; the element isn't on every page. Only add the listener when the element exists:

// app.js
const button = document.querySelector("#color-btn");
if (button) {
  button.addEventListener("click", () => {
    document.body.style.background = "lightyellow";
  });
}

Or, shorter, with the ?. operator, which skips the call when the value is null:

// app.js
button?.addEventListener("click", () => {
  document.body.style.background = "lightyellow";
});

Use this only when the element is really optional. If it should be on the page, a silent ?. hides the real problem: fix the selector or the script's position instead.

Errors that look alike

addEventListener is not a function

getElementsByClassName and querySelectorAll don't return one element, but a list of all matches. A list has no addEventListener, so this fails with a different message:

// app.js
const buttons = document.querySelectorAll(".btn");
buttons.addEventListener("click", () => {});

Uncaught TypeError: buttons.addEventListener is not a function

Add the listener to each element in the list instead:

// app.js
const buttons = document.querySelectorAll(".btn");
buttons.forEach(button => {
  button.addEventListener("click", () => {
    console.log("clicked", button.textContent);
  });
});

Cannot read properties of undefined (reading 'addEventListener')

undefined instead of null usually means you took an item from a list that was empty: document.getElementsByClassName("btn")[0] gives undefined when there's no element with that class. The causes and fixes are the same as above: check the class name and where the script runs.

Cannot set properties of null (setting 'textContent')

If you assign to the element instead of calling a method on it, like document.querySelector("#total").textContent = "5";, Chrome says Cannot set properties of null (setting 'textContent'). It's the same problem: the element wasn't found.

In Firefox and Safari

Other browsers word it differently. Firefox says something like button is null, and Safari something like null is not an object (evaluating 'button.addEventListener'). Whatever the wording, it means the same thing: the value before the dot is null.

If the problem is the opposite one, your styles not showing up instead of your script failing, see CSS not working? 7 reasons your styles don't apply.

How to find the culprit quickly

1. Open the console (F12 in most browsers, then the Console tab) and click the file name and line number next to the error. It takes you to the line that failed.
2. Look at the variable before the dot, here button, and find the line that created it.
3. Log it: console.log(button);. If the console shows null, the selector found nothing.
4. Compare the selector with the HTML: the # or ., the spelling, and whether the element is above or below your script.

The same pattern explains its cousins: (reading 'textContent'), (reading 'value') or (reading 'appendChild') all mean that the element before the dot is null.

The same three questions (what went wrong, where, and what it asks about your code) work for errors in other languages too: How to Read an Error Message: A 3-Step Method for Python, Java, JavaScript and SQL.

We checked every example in this article in Chrome, which writes the messages exactly as shown. If you're just starting with JavaScript, our guide Which programming language should you learn first? explains where it fits next to HTML and CSS.

In our Web course, this error is one of the first things you meet when your JavaScript starts talking to the page, and the AI tutor helps you read the console instead of guessing. You can start with the free Web trial, "Your Card on the Web": one 20-minute session where you build a profile card with HTML and CSS, with nothing to install: craftstackacademy.com/trial.

This article was first published on the Craftstack Academy blog.

Don't. The error message is the most useful thing on your screen. It was written by the people who built the language, and it usually tells you what went wrong and where. You just need to know where to look.

Here's a 3-step method that works the same way in Python, Java, JavaScript and SQL, with the most common beginner errors in each language as examples.

The 3-step method

1. Read the last line first. That's the error type and the message: the what.
2. Find the line number. That's the where. Look at that line, and if nothing is wrong there, look at the line just above it.
3. Translate the message into a question about your code. "unsupported operand types int and str" becomes: "Which value on this line is text when I expected a number?"

That's it. Let's try it on real errors.

Python: the error is at the bottom

Here's a tiny program:

name = input("Your name? ")
year = input("Born in? ")
age = 2026 - year
print("Hi", name, "- you are", age)

Python answers with this:

Traceback (most recent call last):
  File "life.py", line 3, in <module>
    age = 2026 - year
          ~~~~~^~~~~~
TypeError: unsupported operand type(s) for -: 'int' and 'str'

Step 1, the last line: TypeError: unsupported operand type(s) for -: 'int' and 'str'. You tried to subtract, and one side is a number (int) while the other is text (str).

Step 2, the line number: line 3, and newer Python versions even underline the exact spot with ^.

Step 3, the question: which value is text? year, because input() always gives you back text, even when someone types 1990. Convert it first: int(year).

👉 Full explanation, plus the other ways this error appears: Python TypeError: unsupported operand type(s) for 'int' and 'str'

Python also has errors that stop the program before it even starts, like this one:

  File "loop.py", line 3
    print(s)
    ^^^^^
IndentationError: expected an indented block after 'for' statement on line 2

Same method: the last line says what (Python expected an indented block), and the message even tells you where the problem started: the for on line 2. The line after it needs to be indented.

👉 Python IndentationError: expected an indented block after 'if', 'for' or a function

Java: the compiler stops you before you run

Java checks your code before it runs, so many errors show up at compile time:

Main.java:6: error: incompatible types: String cannot be converted to int
        int age = scanner.nextLine();
                                  ^

Step 1: incompatible types: String cannot be converted to int. You're putting text into a box made for whole numbers.

Step 2: Main.java:6 means file Main.java, line 6. The ^ points at the value that doesn't fit.

Step 3: where does this value come from, and what type is it? nextLine() returns a String. Convert it with Integer.parseInt(...), or read a number directly.

👉 Java: incompatible types: String cannot be converted to int

Java's most famous runtime error works the same way. Newer Java versions spell out exactly what was null:

Exception in thread "main" java.lang.NullPointerException: Cannot invoke "String.length()" because "..." is null

Read it as a sentence: I couldn't call length(), because this variable was null. The question to ask: where should this variable have been given a value, and why wasn't it?

👉 Java NullPointerException: Cannot invoke ... because ... is null

JavaScript: open the browser console

In the browser, errors don't pop up on the page. Press F12 (or right-click → Inspect) and open the Console tab. One of the most common messages you'll find there:

Uncaught TypeError: Cannot read properties of null (reading 'addEventListener')

Step 1: you tried to use .addEventListener on something that is null, meaning nothing.

Step 2: the console shows the file and line on the right. Click it to jump straight there.

Step 3: why is this element null? Usually document.querySelector(...) didn't find anything: the selector has a typo, or the script runs before the HTML element exists.

👉 JavaScript: Cannot read properties of null (reading 'addEventListener')

SQL: the last word is your clue

Database errors are short, but they still follow the pattern:

SELECT name FROM visitors WHERE room = Vault;

Error: no such column: Vault

Step 1: SQLite looked for a column called Vault.

Step 2: there's only one line here, so look at the word the message names.

Step 3: is Vault meant to be a column or a value? It's a value, the name of a room, so it needs single quotes: WHERE room = 'Vault'.

👉 SQLite error "no such column": usually missing quotes, plus 4 other causes

When you start counting and grouping rows, you'll meet a different kind of message, about aggregates and GROUP BY. The same 3 steps work:

👉 SQL GROUP BY errors: "misuse of aggregate" and "must appear in the GROUP BY clause"

And when there's no error at all?

The hardest bugs are the silent ones. CSS is the classic example: if a style doesn't apply, the browser doesn't complain at all. Then the method changes slightly: instead of reading a message, you inspect the element (F12 → Elements) and check which styles are crossed out and why.

👉 CSS not working? 7 reasons your styles don't apply

The same goes for code that runs but gives the wrong answer. No message will save you there. You compare what you expected with what you got, and work backwards.

Cheat sheet

Practise reading errors, not just fixing them

The goal isn't to memorise every error. It's to stop being afraid of them. Next time something breaks, before you change a single character, say out loud: what went wrong, where, and what question that raises about your code.

If you're still choosing a first language, start here: Which programming language should you learn first? Python, JavaScript, SQL or Java

Want to practise this with real code? At Craftstack Academy, an AI tutor reads your actual code and your actual error, then gives you a hint about where to look, not the answer. Try one free 20-minute session, with nothing to install: craftstackacademy.com/trial.