Try

if (!window.console) console = ...

未定义的变量不能直接引用。但是，所有全局变量都是全局上下文(浏览器为窗口)同名的属性，访问未定义的属性是可以的。

或者使用if (typeof console === 'undefined') console =…如果你想避开神奇的变量窗口，请参阅@Tim Down的回答。

Try

if (!window.console) console = ...

未定义的变量不能直接引用。但是，所有全局变量都是全局上下文(浏览器为窗口)同名的属性，访问未定义的属性是可以的。

或者使用if (typeof console === 'undefined') console =…如果你想避开神奇的变量窗口，请参阅@Tim Down的回答。

在TypeScript中控制台的存根:

if (!window.console) {
console = {
    assert: () => undefined,
    clear: () => undefined,
    count: () => undefined,
    debug: () => undefined,
    dir: () => undefined,
    dirxml: () => undefined,
    error: () => undefined,
    group: () => undefined,
    groupCollapsed: () => undefined,
    groupEnd: () => undefined,
    info: () => undefined,
    log: () => undefined,
    msIsIndependentlyComposed: (e: Element) => false,
    profile: () => undefined,
    profileEnd: () => undefined,
    select: () => undefined,
    time: () => undefined,
    timeEnd: () => undefined,
    trace: () => undefined,
    warn: () => undefined,
    }
};

我只使用console.log在我的代码。所以我包括一个很短的2眼线

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

注意到OP在IE中使用Firebug，所以假设它是Firebug Lite。当调试器窗口打开时，在IE中定义控制台，这是一种奇怪的情况，但是当Firebug已经在运行时会发生什么呢?不确定，但在这种情况下，"firebugx.js"方法可能是一个很好的测试方法:

来源:

https://code.google.com/p/fbug/source/browse/branches/firebug1.2/lite/firebugx.js?r=187

    if (!window.console || !console.firebug) {
        var names = [
            "log", "debug", "info", "warn", "error", "assert",
            "dir","dirxml","group","groupEnd","time","timeEnd",
            "count","trace","profile","profileEnd"
        ];
        window.console = {};
        for (var i = 0; i < names.length; ++i)
            window.console[names[i]] = function() {}
    }

(更新链接12/2014)

if (typeof console == "undefined") {
  this.console = {
    log: function() {},
    info: function() {},
    error: function() {},
    warn: function() {}
  };
}