JavaScript Debugging
Many beginners quit because they cannot debug.
Debugging for Beginners
Many beginners quit because they cannot debug.
This page shows you how to find out why code does not work.
Debugging means finding and fixing mistakes (bugs) in your code.
Bugs are normal. The skill is learning how to locate them quickly.
Why Code Does Not Work?
Programming code might contain syntax errors, or logical errors.
Many of these errors are difficult to diagnose.
When code fails, beginners often guess what is wrong.
Debugging is the opposite: you check facts.
A good debugging habit is: Read → Reproduce → Reduce → Fix
- Read the error
- Reproduce the problem
- Reduce to a small example
Often, when programming code contains errors, nothing will happen. There are no error messages, and you will get no indications where to search for errors.
Searching for (and fixing) errors in programming code is called code debugging.
JavaScript Debuggers
Debugging is not easy. But fortunately, all modern browsers have a built-in JavaScript debugger.
Built-in debuggers can be turned on and off, forcing errors to be reported to the user.
With a debugger, you can also set breakpoints (places where code execution can be stopped), and examine variables while the code is executing.
Step 1: Look in the Console
The browser console shows errors and messages from JavaScript.
If your code "does nothing", the console often tells you why.
Normally (otherwise follow the steps at the bottom of this page), you activate debugging in your browser with the F12 key, and select Console in the debugger menu.
If you do only one thing: always check the console when something fails.
Step 2: Use console.log()
In JavaScript console.log() prints values to the console.
This helps you to see what your code is doing.
Examples
Example Code
html**
# My First Web Page
console.log("Hello!");Step 3: Confirm Your Assumptions
Many bugs happen because you assume a value is something, but it is not.
Check the value.
Check the type.
Example
Example Code
javascriptlet x = 5;
let y = "5";
console.log(x + y); // 55 (string!)
console.log(x + Number(y)); // 10 (number)Reading Error Messages (Beginner Friendly)
Error messages look scary, but they usually mean one of a few common things.
ReferenceError
Means: This name does not exist.
Often a misspelling or variable not declared.
Example
Example Code
javascriptconsole.log(myValue);
// ReferenceError: myValue is not definedTypeError
Means: You tried to use a value in an impossible way.
Often undefined or null.
Example
Example Code
javascriptlet x;
console.log(x.length);
// TypeError: Cannot read properties of undefinedA Simple Debugging Checklist
- Check the console for errors
- Read the error message carefully
- Log values with console.log()
- Reduce the problem to a small example
- Fix one thing at a time.
Common Beginner Mistakes
Examples
Example Code
javascriptlet x = 10;
if (x = 5) {
console.log("This runs");
}If your if statement behaves strangely, check if you accidentally used =.
Major Browsers' Debugging Tools
You can activate debugging in your browser with F12, and select Console in the debugger menu.
Otherwise follow these steps:
Chrome
- Open the browser.
- From the menu, select "More tools".
- From tools, choose "Developer tools".
- Finally, select Console.
Firefox
- Open the browser.
- From the menu, select "Web Developer".
- Finally, select "Web Console".
Edge
- Open the browser.
- From the menu, select "Developer Tools".
- Finally, select "Console".
Opera
- Open the browser.
- From the menu, select "Developer".
- From "Developer", select "Developer tools".
- Finally, select "Console".
Safari
- Go to Safari, Preferences, Advanced in the main menu.
- Check "Enable Show Develop menu in menu bar".
- When the new option "Develop" appears in the menu:Choose "Show Error Console".
Example
Example Code
htmllet a = 5;
let b = 6;
let c = a + b;
console.log(c);Interactive Code Examples
Examples
htmlJS Example 2
javascriptJS Example 3
javascriptJS Example 4
javascriptExamples
javascriptJS Example 6
html<script>
let a = 5;
let b = 6;
let c = a + b;
console.log(c);