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

当使用React时,你可以利用钩子来管理它,这样它就被限制在你的组件范围内

import { useEffect, useRef } from "react";

type LoggingReplacements = {
  debug?: typeof console["debug"];
  error?: typeof console["error"];
  info?: typeof console["info"];
  log?: typeof console["log"];
  warn?: typeof console["warn"];
};
/**
 * This replaces console.XXX loggers with custom implementations.  It will restore console log on unmount of the component.
 * @param replacements a map of replacement loggers.  They're all optional and only the functions defined will be replaced.
 */
export function useReplaceLogging({
  debug,
  error,
  info,
  log,
  warn,
}: LoggingReplacements): void {
  const originalConsoleDebug = useRef(console.debug);
  const originalConsoleError = useRef(console.error);
  const originalConsoleInfo = useRef(console.info);
  const originalConsoleLog = useRef(console.log);
  const originalConsoleWarn = useRef(console.warn);
  if (debug) {
    console.debug = debug;
  }
  if (error) {
    console.error = error;
  }
  if (info) {
    console.info = info;
  }
  if (log) {
    console.log = log;
  }
  if (warn) {
    console.warn = warn;
  }
  useEffect(() => {
    return function restoreConsoleLog() {
      console.debug = originalConsoleDebug.current;
      console.error = originalConsoleError.current;
      console.info = originalConsoleInfo.current;
      console.log = originalConsoleLog.current;
      console.warn = originalConsoleWarn.current;
    };
  }, []);
}

在https://github.com/trajano/react-hooks/tree/master/src/useReplaceLogging上编写测试代码

其他回答

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?

我在这个url中找到了一段更高级的代码。

var DEBUG_MODE = true; // Set this value to false for production

if(typeof(console) === 'undefined') {
   console = {}
}

if(!DEBUG_MODE || typeof(console.log) === 'undefined') {
   // FYI: Firebug might get cranky...
   console.log = console.error = console.info = console.debug = console.warn = console.trace = console.dir = console.dirxml = console.group = console.groupEnd = console.time =    console.timeEnd = console.assert = console.profile = function() {};
}

禁用console.log:

console.log = function() {};

禁用所有写入控制台的功能。

for (let func in console) {
   console[func] = function() {};
}

在对这个问题做了一些研究和开发之后,我遇到了这个解决方案,它将根据您的选择隐藏警告/错误/日志。

    (function () {
    var origOpen = XMLHttpRequest.prototype.open;
    XMLHttpRequest.prototype.open = function () {        
        console.warn = function () { };
        window['console']['warn'] = function () { };
        this.addEventListener('load', function () {                        
            console.warn('Something bad happened.');
            window['console']['warn'] = function () { };
        });        
    };
})();

将此代码添加到JQuery插件(例如/../ JQuery. min.js)之前,即使这是不需要JQuery的JavaScript代码。因为有些警告是JQuery本身的。

谢谢! !

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。