是否有任何方法关闭我的JavaScript代码中的所有console.log语句,用于测试目的?
当前回答
你可以使用javascript AOP(例如jquery-aop)来拦截所有对console.debug/log的调用,如果某个全局变量被设置为false,就不进行实际的调用。
您甚至可以(不时地)执行ajax调用,这样就可以更改服务器上启用/禁用日志的行为,当在登台环境或类似环境中遇到问题时启用调试,这可能非常有趣。
其他回答
只需更改标志DEBUG以覆盖console.log函数。这应该能奏效。
var DEBUG = false;
// ENABLE/DISABLE Console Logs
if(!DEBUG){
console.log = function() {}
}
我认为2020年最简单、最容易理解的方法是创建一个像log()这样的全局函数,你可以选择以下方法之一:
const debugging = true;
function log(toLog) {
if (debugging) {
console.log(toLog);
}
}
function log(toLog) {
if (true) { // You could manually change it (Annoying, though)
console.log(toLog);
}
}
你可以说这些功能的缺点是:
您仍然在运行时调用函数 您必须记住在第二个选项中更改调试变量或if语句 您需要确保在加载所有其他文件之前加载了该函数
And my retorts to these statements is that this is the only method that won't completely remove the console or console.log function which I think is bad programming because other developers who are working on the website would have to realize that you ignorantly removed them. Also, you can't edit JavaScript source code in JavaScript, so if you really want something to just wipe all of those from the code you could use a minifier that minifies your code and removes all console.logs. Now, the choice is yours, what will you do?
在脚本中重新定义console.log函数。
console.log = function() {}
够了,不再给控制台发消息了。
编辑:
扩展了Cide的想法。一个自定义记录器,您可以使用它从代码中切换登录。
从我的Firefox控制台:
var logger = function()
{
var oldConsoleLog = null;
var pub = {};
pub.enableLogger = function enableLogger()
{
if(oldConsoleLog == null)
return;
window['console']['log'] = oldConsoleLog;
};
pub.disableLogger = function disableLogger()
{
oldConsoleLog = console.log;
window['console']['log'] = function() {};
};
return pub;
}();
$(document).ready(
function()
{
console.log('hello');
logger.disableLogger();
console.log('hi', 'hiya');
console.log('this wont show up in console');
logger.enableLogger();
console.log('This will show up!');
}
);
如何使用上面的“记录器”?在就绪事件中,调用记录器。disableLogger使控制台消息不被记录。向记录器添加调用。enabllogger和logger。在希望将消息记录到控制台的方法中的disableLogger。
我这样写道:
//Make a copy of the old console.
var oldConsole = Object.assign({}, console);
//This function redefine the caller with the original one. (well, at least i expect this to work in chrome, not tested in others)
function setEnabled(bool) {
if (bool) {
//Rewrites the disable function with the original one.
console[this.name] = oldConsole[this.name];
//Make sure the setEnable will be callable from original one.
console[this.name].setEnabled = setEnabled;
} else {
//Rewrites the original.
var fn = function () {/*function disabled, to enable call console.fn.setEnabled(true)*/};
//Defines the name, to remember.
Object.defineProperty(fn, "name", {value: this.name});
//replace the original with the empty one.
console[this.name] = fn;
//set the enable function
console[this.name].setEnabled = setEnabled
}
}
不幸的是,它在使用严格模式下不起作用。
使用console。fn。setEnabled = setEnabled然后是console。fn。setEnabled(false) fn可以是几乎任何控制台函数。 你的情况是:
console.log.setEnabled = setEnabled;
console.log.setEnabled(false);
我还写了这个:
var FLAGS = {};
FLAGS.DEBUG = true;
FLAGS.INFO = false;
FLAGS.LOG = false;
//Adding dir, table, or other would put the setEnabled on the respective console functions.
function makeThemSwitchable(opt) {
var keysArr = Object.keys(opt);
//its better use this type of for.
for (var x = 0; x < keysArr.length; x++) {
var key = keysArr[x];
var lowerKey = key.toLowerCase();
//Only if the key exists
if (console[lowerKey]) {
//define the function
console[lowerKey].setEnabled = setEnabled;
//Make it enabled/disabled by key.
console[lowerKey].setEnabled(opt[key]);
}
}
}
//Put the set enabled function on the original console using the defined flags and set them.
makeThemSwitchable(FLAGS);
所以你只需要在FLAGS中加入默认值(在执行上面的代码之前),比如FLAGS. log = false,日志功能将在默认情况下被禁用,仍然可以调用console.log.setEnabled(true)来启用它
这是在JS 2020中引入的。在浏览器上globalThis和window一样,在nodejs上globalThis和global一样等等。在任何环境上,它将直接指向全局对象,因此这段代码将在任何支持JS2020的env上工作了解更多信息:https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/globalThis
对于任何现代浏览器& nodejs v12或更新版本,你应该使用这个:
globalThis.console.log = () => null;
globalThis.console.warn = () => null;
globalThis.console.info = () => null;
globalThis.console.error = () => null;