10

I'd like to make my JS code production ready by stripping out all console.log("blah blah") debugging statements. I'm confused by this popular SO answer (code below) on how to do this using Google's closure compiler, a popular JS minifier/compiler.

/** @const */
var LOG = false;
...
LOG && log('hello world !'); // compiler will remove this line
...

//this will even work with `SIMPLE_OPTIMALIZATIONS` and no `--define=` is necessary !

Two questions:

  1. Multiples files: How does the above code work with multiple files (basic example below)? It has to be in a closure, right? Doesn't this then mean you have to put this code on each page? Also, doesn't it also then mean you have to change all the variables you want to be global from var foo='bar' to var window.foo='bar'; inside these closures on each page?

  2. Minor issue: Shouldn't it be console.log('...') rather than log('...') because log() gives an error? Am I missing something obvious here?

<script src='/assets/js/file1.js'></script> <script src='/assets/js/file2.js'></script>

contents of file1.js:

var foo='bar';
console.log("foo"+foo);

contents of file2.js

var baz='bazzy';
console.log("baz"+baz);
Improve this question
2
  • log is a generalized logging function which could simply be a wrapper for console.log or could provide some other type of logging. It was simply used for demonstration. The definition is left as an exercise to the reader. Commented Sep 16, 2013 at 21:31
  • @ChadKillingsworth thanks, I didn't know whether it was an exercise or something I hadn't learned yet about JS. Commented Sep 16, 2013 at 21:37

3 Answers 3

Reset to default
2

If you have your code split up into multiple files, I would think you'd want to have a build process that joins them into one immediately invoked function. Then your var LOG = false; can just be included once at the top.

If by "each page", you mean that you have separate JavaScript files per page that aren't meant to be joined together, then yes, you'd need to have that code at the top of each, but then you're also not taking as much advantage of Closure Compiler.

Regarding globals, yes, you'd need to use window when setting, though I would hope you're defining only one global.

The use of log() implies that someone defined a log() function so that it's less verbose.

function log() {
    return console.log.apply(console, arguments);
}
Improve this answer
Sign up to request clarification or add additional context in comments.

6 Comments

thanks! I've now better explained what I mean by "multiple files" in my question. Please see the edits.
-@user2736012 can you provide a basic example of "a build process that joins them into one immediately invoked function"? I'm not quite sure what you mean by that. Closure compiler is making one big JS file from my file1.js, file2.js, jquery.js, etc.
Some server-side code to facilitate an automated process of taking your individual files, joining them together into one file, running desired processes against the file (like Closure Compiler), and outputting the result to a new file to be included with your project.
ok thanks now that's clear. last question: why only one global? just cause good code practice? prevent name collisions? other reasons?
@timpeterson: Yeah, it's considered good practice to minimize the number of globals created mainly for the purpose you described... name collisions.
|
0

I think you don't understand the answer from your link. Author suggests you to create a new variable 

var LOG = false;

then you simply write LOG && __code__ ; ( coressponds to if(LOG) __code__ ; ) and the code after && will not be processed, because of partial evaluation of boolean expressions. It will be processed, after you set LOG to true. The variable is usually called a "switch" or "flag".

By the way, Closure Compiler may edit your code, but that code must behave the same after "compilation" as before it. That is the main principle of all these minifiers and optimizers (they must not convert your Quick Sort into GTA 5). So don't use Closure Compiler to make your code work differently (it is not possible, if it is - they have a bug in it!).

Improve this answer

1 Comment

-@IvanKuckir thanks. "I think you don't understand the answer from your link". Yes that is why I asked the question. :)
0

Taking the OP's example, if you don't want to modify file1.js you can compile it with an additional file (noconsole.js) containing:

console.log = console.debug = console.info = function(){};

Such as:

google-closure-compiler --js noconsole.js file1.js --js_output_file file1.out.js

As explained in my previous answer.

Improve this answer

Comments

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.