Override console.log(); for production [duplicate]

This question already has answers here: How to quickly and conveniently disable all console.log statements in my code? (38 answers) Closed 7 years ago.

I'm fairly new to 开发者_运维技巧Javascript development so this might be a real newbie question.

I've got a sencha-touch application riddled with console.log(); for debugging purposes.

I've got chirpy doing all of my build time combining. It outputs a app.debug.js for debugging as well as a app.min.js for production

Now I could go through all of my code files looking for console.log(); and delete it manually when I'm ready to go to production, but I'm wondering if there's a way to override the method.

Basically, whenever the console.log(); method is called, DO NOTHING.

That way, I can put the override code file in my production config, and NOT in my debug config.

Is this possible?

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Put this at the top of the file:

var console = {};
console.log = function(){};

For some browsers and minifiers, you may need to apply this onto the window object.

window.console = console;

---

Or if you just want to redefine the behavior of the console (in order to add logs for example) You can do something like that:

// define a new console
var console=(function(oldCons){
    return {
        log: function(text){
            oldCons.log(text);
            // Your code
        },
        info: function (text) {
            oldCons.info(text);
            // Your code
        },
        warn: function (text) {
            oldCons.warn(text);
            // Your code
        },
        error: function (text) {
            oldCons.error(text);
            // Your code
        }
    };
}(window.console));

//Then redefine the old console
window.console = console;

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It would be super useful to be able to toggle logging in the production build. The code below turns the logger off by default.

When I need to see logs, I just type debug(true) into the console.

var consoleHolder = console;
function debug(bool){
    if(!bool){
        consoleHolder = console;
        console = {};
        Object.keys(consoleHolder).forEach(function(key){
            console[key] = function(){};
        })
    }else{
        console = consoleHolder;
    }
}
debug(false);

To be thorough, this overrides ALL of the console methods, not just console.log.

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I use something similar to what posit labs does. Save the console in a closure and you have it all in one portable function.

var GlobalDebug = (function () {
    var savedConsole = console;
    return function(debugOn,suppressAll){
        var suppress = suppressAll || false;
        if (debugOn === false) {
            console = {};
            console.log = function () { };
            if(suppress) {
                console.info = function () { };
                console.warn = function () { };
                console.error = function () { };
            } else {
                console.info = savedConsole.info;
                console.warn = savedConsole.warn;
                console.error = savedConsole.error;              
            }
        } else {
            console = savedConsole;
        }
    }
})();

Just do globalDebug(false) to toggle log messages off or globalDebug(false,true) to remove all console messages.

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console.log = function(){};

Override it like any other thing.

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I would recommend using: https://github.com/sunnykgupta/jsLogger

Features:

1. It safely overrides the console.log.
2. Takes care if the console is not available (oh yes, you need to factor that too.)
3. Stores all logs (even if they are suppressed) for later retrieval.
4. Handles major console functions like log, warn, error, info.

Is open for modifications and will be updated whenever new suggestions come up.

Disclaimer: I am the author of the plugin.

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You could also use regex to delete all the console.log() calls in your code if they're no longer required. Any decent IDE will allow you to search and replace these across an entire project, and allow you to preview the matches before committing the change.

\s*console\.log\([^)]+\);

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After read a lot of posts, I made my own solution as follow:

SCRIPT:

function extendConsole() {
    "use strict";
    try {
        var disabledConsoles = {};

        console.enable = function (level, enabled) {
            // Prevent errors in browsers without console[level]
            if (window.console === 'undefined' || !window.console || window.console === null) {
                window.console = {};
            }
            if (window.console[level] === 'undefined' || !window.console[level] || window.console[level] == null) {
                window.console[level] = function() {};
            }

            if (enabled) {
                if (disabledConsoles[level]) {
                    window.console[level] = disabledConsoles[level];
                }
                console.info("console." + level + "() was enabled.");
            } else {
                disabledConsoles[level] = window.console[level];
                window.console[level] = function () { };
                console.info("console." + level + "() was disabled.");
            }
        };
    } catch (exception) {
        console.error("extendConsole() threw an exception.");
        console.debug(exception);
    }
}

USAGE:

extendConsole();
console.enable("debug", window.debugMode);

EXAMPLE:

http://jsfiddle.net/rodolphobrock/2rzxb5bo/10/

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This will override console.log function when the url does not contain localhost. You can replace the localhost with your own development settings.

// overriding console.log in production
if(window.location.host.indexOf('localhost:9000') < 0) {
    console.log = function(){};
}

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Just remember that with this method each console.log call will still do a call to a (empty) function causing overhead, if there are 100 console.log commands, you are still doing 100 calls to a blank function.

Not sure how much overhead this would cause, but there will be some, it would be preferable to have a flag to turn debug on then use something along the lines of:

var debug=true; if (debug) console.log('blah')

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Theres no a reason to let all that console.log all over your project in prod enviroment... If you want to do it on the proper way, add UglifyJS2 to your deployment process using "drop_console" option.

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Here is what I did

    var domainNames =["fiddle.jshell.net"]; // we replace this by our production domain.

var logger = {
    force:false,
    original:null,
    log:function(obj)
    {
        var hostName = window.location.hostname;
        if(domainNames.indexOf(hostName) > -1)
        {
            if(window.myLogger.force === true)
            {
                window.myLogger.original.apply(this,arguments);
            }
        }else {
            window.myLogger.original.apply(this,arguments);
        }
    },
    forceLogging:function(force){
        window.myLogger.force = force;
    },
    original:function(){
        return window.myLogger.original;
    },
    init:function(){
        window.myLogger.original = console.log;
        console.log = window.myLogger.log;
    }
}

window.myLogger = logger;
console.log("this should print like normal");
window.myLogger.init();
console.log("this should not print");
window.myLogger.forceLogging(true);
console.log("this should print now");

Also posted about it here. http://bhavinsurela.com/naive-way-of-overriding-console-log/

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You can look into UglifyJS: http://jstarrdewar.com/blog/2013/02/28/use-uglify-to-automatically-strip-debug-messages-from-your-javascript/, https://github.com/mishoo/UglifyJS I haven't tried it yet.

Quoting,

 if (typeof DEBUG === 'undefined') DEBUG = true; // will be removed

 function doSomethingCool() {
     DEBUG && console.log("something cool just happened"); // will be removed }

...The log message line will be removed by Uglify's dead-code remover (since it will erase any conditional that will always evaluate to false). So will that first conditional. But when you are testing as uncompressed code, DEBUG will start out undefined, the first conditional will set it to true, and all your console.log() messages will work.