We're yet to see a JavaScript developer that doesn't use console.log
to perform a "quick debug". And it goes like this:
let theAnswer = 42
console.log(theAnswer);
And the console will oblidge, showing:
42
Encapsulate in an object
Here is a handy tip: encapsulate the variable in an object and the variable name will be displayed:
console.log( { theAnswer } );
The output will now be:
{theAnswer: 42}
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