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
如果你使用IE7,控制台将不会被定义。所以一个更IE友好的版本是:
if (typeof console == "undefined" || typeof console.log == "undefined")
{
var console = { log: function() {} };
}
其他回答
我一直在用以下方法来处理这个问题:-
var debug = 1;
var logger = function(a,b){ if ( debug == 1 ) console.log(a, b || "");};
将debug设置为1以启用调试。然后在输出调试文本时使用记录器函数。它还设置为接受两个参数。
所以,与其
console.log("my","log");
use
logger("my","log");
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() {};
}
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() {};
}
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
我写了一个ES2015解决方案(仅用于Webpack)。
class logger {
static isEnabled = true;
static enable () {
if(this.constructor.isEnabled === true){ return; }
this.constructor.isEnabled = true;
}
static disable () {
if(this.constructor.isEnabled === false){ return; }
this.constructor.isEnabled = false;
}
static log () {
if(this.constructor.isEnabled === false ) { return; }
const copy = [].slice.call(arguments);
window['console']['log'].apply(this, copy);
}
static warn () {
if(this.constructor.isEnabled === false ) { return; }
const copy = [].slice.call(arguments);
window['console']['warn'].apply(this, copy);
}
static error () {
if(this.constructor.isEnabled === false ) { return; }
const copy = [].slice.call(arguments);
window['console']['error'].apply(this, copy);
}
}
描述:
Along with logger.enable and logger.disable you can use console.['log','warn','error'] methods as well using logger class. By using logger class for displaying, enabling or disabling messages makes the code much cleaner and maintainable. The code below shows you how to use the logger class: logger.disable() - disable all console messages logger.enable() - enable all console messages logger.log('message1', 'message2') - works exactly like console.log. logger.warn('message1', 'message2') - works exactly like console.warn. logger.error('message1', 'message2') - works exactly like console.error. Happy coding..