JavaScript - 如何获取调用脚本的URL？ [英] JavaScript - How do I get the URL of script being called?

问题描述

我在文件 http://site1.com/index.html 中包含myscript.js，如下所示：

I include myscript.js in the file http://site1.com/index.html like this:

<script src=http://site2.com/myscript.js></script>

在myscript.js中，我希望能够访问URL http://site2.com/myscript.js 。我想要这样的东西：

Inside "myscript.js", I want to get access to the URL "http://site2.com/myscript.js". I'd like to have something like this:

function getScriptURL() {
    // something here
    return s
}

alert(getScriptURL());

哪会警告 http://site2.com/myscript.js 如果从上面提到的index.html调用。

Which would alert "http://site2.com/myscript.js" if called from the index.html mentioned above.

推荐答案

来自 http://feather.elektrum.org/book/src.html ：

var scripts = document.getElementsByTagName('script');
var index = scripts.length - 1;
var myScript = scripts[index];

变量 myScript 现在有脚本dom元件。您可以使用 myScript.src 获取src网址。

The variable myScript now has the script dom element. You can get the src url by using myScript.src.

请注意，这需要作为初步评估脚本。如果您不想污染Javascript命名空间，您可以执行以下操作：

Note that this needs to execute as part of the initial evaluation of the script. If you want to not pollute the Javascript namespace you can do something like:

var getScriptURL = (function() {
    var scripts = document.getElementsByTagName('script');
    var index = scripts.length - 1;
    var myScript = scripts[index];
    return function() { return myScript.src; };
})();

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