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
https://stackoverflow.com/a/46189791/871166的简化
switch (process.env.LOG_LEVEL) {
case 'ERROR':
console.warn = function() {};
case 'WARN':
console.info = function() {};
case 'INFO':
console.log = function() {};
case 'LOG':
console.debug = function() {};
console.dir = function() {};
}
其他回答
我自己弄明白后发现了这个帖子。以下是我的解决方案:
const testArray = {
a: 1,
b: 2
};
const verbose = true; //change this to false to turn off all comments
const consoleLog = (...message) => {
return verbose ? console.log(...message) : null;
};
console.log("from console.log", testArray);
consoleLog("from consoleLog", testArray);
// use consoleLog() for the comments you want to be able to toggle.
在其他答案的基础上,我个人希望只关闭代码的特定部分(ES6模块,但简单的单独脚本也可以)。
// old console to restore functionality
const consoleHolder = window.console;
// arbitrary strings, for which the console stays on (files which you aim to debug)
const debuggedHandlers = ["someScript", "anotherScript"];
// get console methods and create a dummy with all of them empty
const consoleMethodKeys = Object.getOwnPropertyNames(window.console).filter(item => typeof window.console[item] === 'function');
const consoleDummy = {};
consoleMethodKeys.forEach(method => consoleDummy[method] = () => {});
export function enableConsoleRedirect(handler) {
if (!debuggedHandlers.includes(handler)) {
window.console = consoleDummy;
}
}
export function disableConsoleRedirect() {
window.console = consoleHolder;
}
然后,只需将这个模块导入到您希望能够切换调试模式的任何文件中,在文件顶部调用enable函数,在底部调用disable函数。
如果希望在简单脚本中使用它,可能需要将顶部包装在匿名函数中和/或稍微重新组织它,以最大限度地减少名称空间污染。
此外,你可能想要只使用true/false而不是字符串处理程序,并在当前使用的文件中切换调试模式。
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。
这应该覆盖window.console的所有方法。你可以把它放在你的脚本部分的最上面,如果你在一个PHP框架上,你只能在你的应用程序环境是生产的时候打印这段代码,或者当某种调试标志被禁用的时候。然后,代码中的所有日志都将在开发环境或调试模式下工作。
window.console = (function(originalConsole){
var api = {};
var props = Object.keys(originalConsole);
for (var i=0; i<props.length; i++) {
api[props[i]] = function(){};
}
return api;
})(window.console);
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?