如何获取参数名称？ [复制]

Question

我有一个类似的网址：

//my-url-adresses.com/web/Assets/js/javascripts?v=5.2&marker=2017-05/logo_30dKmOHPNLq8CfiOYfqhmarkerv2.png&cdn=//cdn.mywebsite.com/files/

我想得到这部分：（在 cdn= 之后）：

//cdn.mywebsite.com/files/

我在下面提供了我的代码。请纠正我错的地方。提前致谢。

var scripts = document.getElementsByTagName('script');
var index = scripts.length - 1;
var myScript = scripts[index];
var scripturl = myScript.src;
var cdnUrl = getParameterByName("cdn", scripturl);
var markerUrl = getParameterByName("marker", scripturl);


function getParameterByName(name, url) {
    if (!url) url = window.location.href;
    name = name.replace(/[\[\]]/g, "\\$&");
    var regex = new RegExp("[?&]" + name + "(=([^&#]*)|&|#|$)"),
        results = regex.exec(url);
    if (!results) return null;
    if (!results[2]) return '';
    return decodeURIComponent(results[2].replace(/\+/g, " "));
}


console.log(markerUrl);

<script id="scripts" src="//my-url-adresses.com/web/Assets/js/javascripts?v=5.2&marker=2017-05/logo_30dKmOHPNLq8CfiOYfqhmarkerv2.png&cdn=//cdn.mywebsite.com.com/files/" defer></script>

Answer 1

我注意到您的代码无法使用 index 属性访问脚本 url。我还注意到您的脚本标签中有 id 。最好使用它，否则如果您在页面底部的页面中有一些带有某些功能的脚本标签，它会得到那些（因为您在页面中找到最后一个脚本标签）并且因为它没有有src 它不会获取结果。

这是使用getElementById()的解决方案

var myScript = document.getElementById('scripts');
var scripturl = myScript.src;
var cdnUrl = getParameterByName("cdn", scripturl);
var markerUrl = getParameterByName("marker", scripturl);


function getParameterByName(name, url) {
    url = decodeURIComponent(url);
    name = name.replace(/[\[\]]/g, "\\$&");
    var regex = new RegExp("[?&]" + name + "(=([^&#]*)|&|#|$)"),
        results = regex.exec(url);
    if (!results) return null;
    if (!results[2]) return '';
    return decodeURIComponent(results[2].replace(/\+/g, " "));
}


console.log(markerUrl);

<script id="scripts" src="//my-url-adresses.com/web/Assets/js/javascripts?v=5.2&marker=2017-05/logo_30dKmOHPNLq8CfiOYfqhmarkerv2.png&cdn=//cdn.anitur.com/web/" defer></script>

Answer 2

试试这个：

   var myScript = document.getElementById('scripts');
var scripturl = myScript.src; // or window.location.href for current url
    var captured = /cdn=([^&]+)/.exec(scripturl )[1];  
    var result = captured ? captured : 'myDefaultValue';
    console.log(result);