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如何找出javascript中加载了哪些javascript？

发布于 2024-09-30 16:09:37 字数 254 浏览 9 评论 0原文

继另一个问题的评论之后，我问自己是否有办法获取页面上加载的所有 js 代码的列表。就像 Firebug 或 chrome Inspector 所做的那样。

有没有一个纯javascript的方法？

一种方法是抓取脚本标签，但这样你可能会错过动态加载的 js。我希望有一个API。

对于另一个问题，可以 grep 所有调用 console.debug() 的脚本，以防止忘记它们并让它们进入生产环境。

谢谢

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梦忆晨望 2024-10-07 16:09:37

我想不出不需要做很多工作的事情。这是我最初失败的尝试。当它尝试迭代 window 的所有内部属性时，它会陷入无限递归。

/**
 * You have to run this in firefox, pass window the first time
 * @return boolean Whether the given object contains a function where its
 * source code contains the word console.
 */ 
function lookForConsole( obj ) {
  var found  = false;
  for (var prop in obj) {
    var current = obj[prop];
    if (typeof current == "function") {

      if (current.toSource.indexOf("console" + ".log") != -1) {
        found = true;
        break;
      }
    } else if (typeof current == "object"){
      found = lookForConsole(current);
      if (found) {
        break;
      }
    }
  }
  return found;
}

您是否听过这样一句话：“当您拥有的唯一工具是锤子时，所有问题看起来都像钉子”？

为什么要在 JS 中这样做？

I can't think of something that doesn't require a lot of work. Here's my initial failed attempt. It gets into infinite recursion as it tries to iterate through all the inner properties of window.

/**
 * You have to run this in firefox, pass window the first time
 * @return boolean Whether the given object contains a function where its
 * source code contains the word console.
 */ 
function lookForConsole( obj ) {
  var found  = false;
  for (var prop in obj) {
    var current = obj[prop];
    if (typeof current == "function") {

      if (current.toSource.indexOf("console" + ".log") != -1) {
        found = true;
        break;
      }
    } else if (typeof current == "object"){
      found = lookForConsole(current);
      if (found) {
        break;
      }
    }
  }
  return found;
}

You ever hear the expression, "when the only tool you have is a hammer, every problem looks like a nail"?

Why would you do this in JS?

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勿忘初心 2024-10-07 16:09:37

使用 jQuery：

$(document).ready(function() {


$('script').each(function() {
    if($(this).attr('src')) {
        alert($(this).attr('src'))
    }
    else {
        alert("inline")
    }
})

});

using jQuery:

$(document).ready(function() {


$('script').each(function() {
    if($(this).attr('src')) {
        alert($(this).attr('src'))
    }
    else {
        alert("inline")
    }
})

});

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瑶笙 2024-10-07 16:09:37

这就是萤火虫的做法。我想那就没有更好的方法了。

    var doc = Firebug.browser.document;
    var scripts = doc.getElementsByTagName("script");
    var selectNode = this.selectNode = createElement("select");

    for(var i=0, script; script=scripts[i]; i++)
    {
        // Don't show Firebug Lite source code in the list of options
        if (Firebug.ignoreFirebugElements && script.getAttribute("firebugIgnore"))
            continue;

        var fileName = getFileName(script.src) || getFileName(doc.location.href);
        var option = createElement("option", {value:i});

        option.appendChild(Firebug.chrome.document.createTextNode(fileName));
        selectNode.appendChild(option);
    };

http://fbug.googlecode.com/svn /lite/branches/firebug1.3/content/firebug/script.js

This is how firebug does it. I guess there's no fancier way then.

    var doc = Firebug.browser.document;
    var scripts = doc.getElementsByTagName("script");
    var selectNode = this.selectNode = createElement("select");

    for(var i=0, script; script=scripts[i]; i++)
    {
        // Don't show Firebug Lite source code in the list of options
        if (Firebug.ignoreFirebugElements && script.getAttribute("firebugIgnore"))
            continue;

        var fileName = getFileName(script.src) || getFileName(doc.location.href);
        var option = createElement("option", {value:i});

        option.appendChild(Firebug.chrome.document.createTextNode(fileName));
        selectNode.appendChild(option);
    };

http://fbug.googlecode.com/svn/lite/branches/firebug1.3/content/firebug/script.js

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~没有更多了~