function getUrlVar(key){
    var result = new RegExp(key + "=([^&]*)", "i").exec(window.location.search); 
    return result && unescape(result[1]) || ""; 
}

https://gist.github.com/1771618

此函数将查询字符串转换为类似JSON的对象，它还处理无值和多值参数：

"use strict";
function getQuerystringData(name) {
    var data = { };
    var parameters = window.location.search.substring(1).split("&");
    for (var i = 0, j = parameters.length; i < j; i++) {
        var parameter = parameters[i].split("=");
        var parameterName = decodeURIComponent(parameter[0]);
        var parameterValue = typeof parameter[1] === "undefined" ? parameter[1] : decodeURIComponent(parameter[1]);
        var dataType = typeof data[parameterName];
        if (dataType === "undefined") {
            data[parameterName] = parameterValue;
        } else if (dataType === "array") {
            data[parameterName].push(parameterValue);
        } else {
            data[parameterName] = [data[parameterName]];
            data[parameterName].push(parameterValue);
        }
    }
    return typeof name === "string" ? data[name] : data;
}

我们对参数[1]执行未定义检查，因为如果变量未定义，decodeURIComponent将返回字符串“undefined”，这是错误的。

用法：

"use strict";
var data = getQuerystringData();
var parameterValue = getQuerystringData("parameterName");

function GetQueryStringParams(sParam)
{
    var sPageURL = window.location.search.substring(1);
    var sURLVariables = sPageURL.split('&');

    for (var i = 0; i < sURLVariables.length; i++)
    {
        var sParameterName = sURLVariables[i].split('=');
        if (sParameterName[0] == sParam)
        {
            return sParameterName[1];
        }
    }
}​

假设URL为

http://example.com/?stringtext=jquery&stringword=jquerybyexample

var tech = GetQueryStringParams('stringtext');
var blog = GetQueryStringParams('stringword');

查看此帖子或使用此：

<script type="text/javascript" language="javascript">
    $(document).ready(function()
    {
        var urlParams = {};
        (function ()
        {
            var match,
            pl= /\+/g,  // Regular expression 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;
        }
    });
</script>

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;
};

此函数将根据需要使用递归返回已解析的JavaScript对象，其中包含任意嵌套的值。

这里有一个jsfiddle示例。

[
  '?a=a',
  '&b=a',
  '&b=b',
  '&c[]=a',
  '&c[]=b',
  '&d[a]=a',
  '&d[a]=x',
  '&e[a][]=a',
  '&e[a][]=b',
  '&f[a][b]=a',
  '&f[a][b]=x',
  '&g[a][b][]=a',
  '&g[a][b][]=b',
  '&h=%2B+%25',
  '&i[aa=b',
  '&i[]=b',
  '&j=',
  '&k',
  '&=l',
  '&abc=foo',
  '&def=%5Basf%5D',
  '&ghi=[j%3Dkl]',
  '&xy%3Dz=5',
  '&foo=b%3Dar',
  '&xy%5Bz=5'
].join('');

给出以上任何测试示例。

var qs = function(a) {
  var b, c, e;
  b = {};
  c = function(d) {
    return d && decodeURIComponent(d.replace(/\+/g, " "));
  };
  e = function(f, g, h) {
    var i, j, k, l;
    h = h ? h : null;
    i = /(.+?)\[(.+?)?\](.+)?/g.exec(g);
    if (i) {
      [j, k, l] = [i[1], i[2], i[3]]
      if (k === void 0) {
        if (f[j] === void 0) {
          f[j] = [];
        }
        f[j].push(h);
      } else {
        if (typeof f[j] !== "object") {
          f[j] = {};
        }
        if (l) {
          e(f[j], k + l, h);
        } else {
          e(f[j], k, h);
        }
      }
    } else {
      if (f.hasOwnProperty(g)) {
        if (Array.isArray(f[g])) {
          f[g].push(h);
        } else {
          f[g] = [].concat.apply([f[g]], [h]);
        }
      } else {
        f[g] = h;
      }
      return f[g];
    }
  };
  a.replace(/^(\?|#)/, "").replace(/([^#&=?]+)?=?([^&=]+)?/g, function(m, n, o) {
    n && e(b, c(n), c(o));
  });
  return b;
};