这是基于Gazoris的答案，但URL解码了参数，因此当它们包含除数字和字母以外的数据时可以使用:

function urlParam(name){
    var results = new RegExp('[\?&]' + name + '=([^&#]*)').exec(window.location.href);
    // Need to decode the URL parameters, including putting in a fix for the plus sign
    // https://stackoverflow.com/a/24417399
    return results ? decodeURIComponent(results[1].replace(/\+/g, '%20')) : null;
}

我希望这能有所帮助。

 <script type="text/javascript">
   function getParameters() {
     var searchString = window.location.search.substring(1),
       params = searchString.split("&"),
       hash = {};

     if (searchString == "") return {};
     for (var i = 0; i < params.length; i++) {
       var val = params[i].split("=");
       hash[unescape(val[0])] = unescape(val[1]);
     }

     return hash;
   }

    $(window).load(function() {
      var param = getParameters();
      if (typeof param.sent !== "undefined") {
        // Do something.
      }
    });
</script>

我希望使用完整的简单REG Exp

  function getQueryString1(param) {
    return decodeURIComponent(
        (location.search.match(RegExp("[?|&]"+param+'=(.+?)(&|$)'))||[,null])[1]
    );
  }

这可能有点过分了，但是现在有一个非常流行的用于解析uri的库，叫做URI.js。

例子

var uri = "http://example.org/foo.html?technology=jquery&technology=css&blog=stackoverflow"; var components = URI.parse(uri); var query = URI.parseQuery(components['query']); document.getElementById("result").innerHTML = "URI = " + uri; document.getElementById("result").innerHTML += "<br>technology = " + query['technology']; // If you look in your console, you will see that this library generates a JS array for multi-valued queries! console.log(query['technology']); console.log(query['blog']); <script src="https://cdnjs.cloudflare.com/ajax/libs/URI.js/1.17.0/URI.min.js"></script> <span id="result"></span>

这是基于Gazoris的答案，但URL解码了参数，因此当它们包含除数字和字母以外的数据时可以使用:

function urlParam(name){
    var results = new RegExp('[\?&]' + name + '=([^&#]*)').exec(window.location.href);
    // Need to decode the URL parameters, including putting in a fix for the plus sign
    // https://stackoverflow.com/a/24417399
    return results ? decodeURIComponent(results[1].replace(/\+/g, '%20')) : null;
}

还有另一种功能……

function param(name) {
    return (location.search.split(name + '=')[1] || '').split('&')[0];
}