- Tags: Error, JS Statement
- Categories: JavaScript Reference, JavaScript Statements, Reference
Quick Reference
The debugger statement stops the execution of a script and calls the debugger. If debugging is unavailable, the statement has no effect.
<!-- html element to place output -->
<p id="my_output"></p>
In the following, if debugging is turned on, the script will NOT execute after line 3.
// variable
let x = 'Maserati';
debugger;
// output to HTML element
document.getElementById('my_output').innerHTML = x;
Output
Since nothing after line 3 will execute, there will be no output to the page.
Syntax
debugger;
We’d like to acknowledge that we learned a great deal of our coding from W3Schools and TutorialsPoint, borrowing heavily from their teaching process and excellent code examples. We highly recommend both sites to deepen your experience, and further your coding journey. We’re just hitting the basics here at 1SMARTchicken.
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JAVASCRIPT NOTES:
- When using JavaScript, single or double quotation marks are acceptable and work identically to one another; choose whichever you prefer, and stay consistent
- JavaScript is a case-sensitive language; firstName is NOT the same as firstname
- Arrays count starting from zero NOT one; so item 1 is position [0], item 2 is position [1], and item 3 is position [2] … and so on
- JavaScript variables must begin with a letter, $, or _
- JavaScript variables are case sensitive (x is not the same as X)
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