JavaScript za pomocą Debuggera
let a = 6;
let b = 9;
let c = a * b;
// stops the execution
debugger;
console.log(c);
SAMER SAEID
let a = 6;
let b = 9;
let c = a * b;
// stops the execution
debugger;
console.log(c);// js debugging exercises
// use const and let instead of var
const val1 = 69; // const for non changing values
let val2 = 420; // let for changing values
// use lots of console logs
console.log(val2, 't1')
/* manipulate val2 here */
console.log(val2, 't2')
// note: add some sort of string while console logging,
// so it becomes easy to find & remove logs when code is working
// use try...catch for catching errors
try {
/* your code here */
} catch (err) {
/* handle errors here */
console.log(err, 'error')
}<!DOCTYPE html>
<html>
<body>
<h1>My First Web Page</h1>
<script>
a = 5;
b = 6;
c = a + b;
console.log(c);
</script>
</body>
</html>console.log('Hello, I am writing to the console.');console.log("here debug your code");