获取查询的一行代码：

var value = location.search.match(new RegExp(key + "=(.*?)($|\&)", "i"))[1];

我认为这是实现这一点的准确和简洁的方法（修改自http://css-tricks.com/snippets/javascript/get-url-variables/):

function getQueryVariable(variable) {

    var query = window.location.search.substring(1),            // Remove the ? from the query string.
        vars = query.split("&");                                // Split all values by ampersand.

    for (var i = 0; i < vars.length; i++) {                     // Loop through them...
        var pair = vars[i].split("=");                          // Split the name from the value.
        if (pair[0] == variable) {                              // Once the requested value is found...
            return ( pair[1] == undefined ) ? null : pair[1];   // Return null if there is no value (no equals sign), otherwise return the value.
        }
    }

    return undefined;                                           // Wasn't found.

}

如果您不想使用JavaScript库，可以使用JavaScript字符串函数来解析window.location。将此代码保存在外部.js文件中，您可以在不同的项目中反复使用它。

// Example - window.location = "index.htm?name=bob";

var value = getParameterValue("name");

alert("name = " + value);

function getParameterValue(param)
{
    var url = window.location;
    var parts = url.split('?');
    var params = parts[1].split('&');
    var val = "";

    for ( var i=0; i<params.length; i++)
    {
        var paramNameVal = params[i].split('=');

        if ( paramNameVal[0] == param )
        {
            val = paramNameVal[1];
        }
    }
    return val;
}

Use:

  $(document).ready(function () {
      var urlParams = {};
      (function () {
          var match,
          pl = /\+/g, // Regex for replacing addition symbol with a space
              search = /([^&=]+)=?([^&]*)/g,
              decode = function (s) {
                  return decodeURIComponent(s.replace(pl, " "));
              },
              query = window.location.search.substring(1);

          while (match = search.exec(query))
              urlParams[decode(match[1])] = decode(match[2]);
      })();
      if (urlParams["q1"] === 1) {
          return 1;
      }

请检查并让我知道您的意见。另请参阅How to get querystring value using jQuery。

最漂亮但最基本的：

data = {};
$.each(
    location.search.substr(1).split('&').filter(Boolean).map(function(kvpairs){
        return kvpairs.split('=')
    }),
    function(i,values) {
        data[values.shift()] = values.join('=')
    }
);

它不处理值列表，例如？a[]=1&a[]2

这是Andy E链接的“句柄数组样式查询字符串”版本的扩展版本。修复了一个错误（？key=1&key[]=2&key[]＝3；1丢失并替换为[2,3]），进行了一些小的性能改进（重新解码值，重新计算“[”位置等），并添加了一些改进（功能化，支持？key=1&key=2，支持；分隔符）。我将变量留得很短，但添加了大量注释以使其可读（哦，我在本地函数中重用了v，如果这令人困惑，很抱歉；）。

它将处理以下查询字符串。。。

?test=Hello&pers=neek&pers[]=jeff&pers[][]=jim&pers[extra]=john&test3&nocache=13989148914891264

…把它做成一个看起来像。。。

{
    "test": "Hello",
    "person": {
        "0": "neek",
        "1": "jeff",
        "2": "jim",
        "length": 3,
        "extra": "john"
    },
    "test3": "",
    "nocache": "1398914891264"
}

如上所述，此版本处理一些“格式错误”数组，即-person=neek&person[]=jeff&person[]=jim或person=neek/person=jeff/person=jim，因为密钥是可识别的和有效的（至少在dotNet的NameValueCollection.Add中）：

如果目标NameValueCollection中已存在指定的键例如，指定的值将添加到现有的逗号分隔的格式为“value1，value2，value3”的值列表。

似乎陪审团对重复的键有点不满意，因为没有规范。在这种情况下，多个键被存储为一个（假）数组。但请注意，我不会将基于逗号的值处理为数组。

代码：

getQueryStringKey = function(key) {
    return getQueryStringAsObject()[key];
};


getQueryStringAsObject = function() {
    var b, cv, e, k, ma, sk, v, r = {},
        d = function (v) { return decodeURIComponent(v).replace(/\+/g, " "); }, //# d(ecode) the v(alue)
        q = window.location.search.substring(1), //# suggested: q = decodeURIComponent(window.location.search.substring(1)),
        s = /([^&;=]+)=?([^&;]*)/g //# original regex that does not allow for ; as a delimiter:   /([^&=]+)=?([^&]*)/g
    ;

    //# ma(make array) out of the v(alue)
    ma = function(v) {
        //# If the passed v(alue) hasn't been setup as an object
        if (typeof v != "object") {
            //# Grab the cv(current value) then setup the v(alue) as an object
            cv = v;
            v = {};
            v.length = 0;

            //# If there was a cv(current value), .push it into the new v(alue)'s array
            //#     NOTE: This may or may not be 100% logical to do... but it's better than loosing the original value
            if (cv) { Array.prototype.push.call(v, cv); }
        }
        return v;
    };

    //# While we still have key-value e(ntries) from the q(uerystring) via the s(earch regex)...
    while (e = s.exec(q)) { //# while((e = s.exec(q)) !== null) {
        //# Collect the open b(racket) location (if any) then set the d(ecoded) v(alue) from the above split key-value e(ntry) 
        b = e[1].indexOf("[");
        v = d(e[2]);

        //# As long as this is NOT a hash[]-style key-value e(ntry)
        if (b < 0) { //# b == "-1"
            //# d(ecode) the simple k(ey)
            k = d(e[1]);

            //# If the k(ey) already exists
            if (r[k]) {
                //# ma(make array) out of the k(ey) then .push the v(alue) into the k(ey)'s array in the r(eturn value)
                r[k] = ma(r[k]);
                Array.prototype.push.call(r[k], v);
            }
            //# Else this is a new k(ey), so just add the k(ey)/v(alue) into the r(eturn value)
            else {
                r[k] = v;
            }
        }
        //# Else we've got ourselves a hash[]-style key-value e(ntry) 
        else {
            //# Collect the d(ecoded) k(ey) and the d(ecoded) sk(sub-key) based on the b(racket) locations
            k = d(e[1].slice(0, b));
            sk = d(e[1].slice(b + 1, e[1].indexOf("]", b)));

            //# ma(make array) out of the k(ey) 
            r[k] = ma(r[k]);

            //# If we have a sk(sub-key), plug the v(alue) into it
            if (sk) { r[k][sk] = v; }
            //# Else .push the v(alue) into the k(ey)'s array
            else { Array.prototype.push.call(r[k], v); }
        }
    }

    //# Return the r(eturn value)
    return r;
};