非常轻量级的jQuery方法：

var qs = window.location.search.replace('?','').split('&'),
    request = {};
$.each(qs, function(i,v) {
    var initial, pair = v.split('=');
    if(initial = request[pair[0]]){
        if(!$.isArray(initial)) {
            request[pair[0]] = [initial]
        }
        request[pair[0]].push(pair[1]);
    } else {
        request[pair[0]] = pair[1];
    }
    return;
});
console.log(request);

例如，提醒？q

alert(request.q)

function GET() {
        var data = [];
        for(x = 0; x < arguments.length; ++x)
            data.push(location.href.match(new RegExp("/\?".concat(arguments[x],"=","([^\n&]*)")))[1])
                return data;
    }


example:
data = GET("id","name","foo");
query string : ?id=3&name=jet&foo=b
returns:
    data[0] // 3
    data[1] // jet
    data[2] // b
or
    alert(GET("id")[0]) // return 3

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.

}

下面是String原型实现：

String.prototype.getParam = function( str ){
    str = str.replace(/[\[]/,"\\\[").replace(/[\]]/,"\\\]");
    var regex = new RegExp( "[\\?&]*"+str+"=([^&#]*)" );    
    var results = regex.exec( this );
    if( results == null ){
        return "";
    } else {
        return results[1];
    }
}

示例调用：

var status = str.getParam("status")

str可以是查询字符串或url

Artem Barger回答的改进版本：

function getParameterByName(name) {
    var match = RegExp('[?&]' + name + '=([^&]*)').exec(window.location.search);
    return match && decodeURIComponent(match[1].replace(/\+/g, ' '));
}

有关改进的更多信息，请参阅：http://james.padolsey.com/javascript/bujs-1-getparameterbyname/

tl；博士

一个快速、完整的解决方案，可处理多值键和编码字符。

// using ES5   (200 characters)
var qd = {};
if (location.search) location.search.substr(1).split("&").forEach(function(item) {var s = item.split("="), k = s[0], v = s[1] && decodeURIComponent(s[1]); (qd[k] = qd[k] || []).push(v)})

// using ES6   (23 characters cooler)
var qd = {};
if (location.search) location.search.substr(1).split`&`.forEach(item => {let [k,v] = item.split`=`; v = v && decodeURIComponent(v); (qd[k] = qd[k] || []).push(v)})

// as a function with reduce
function getQueryParams() {
  return location.search
    ? location.search.substr(1).split`&`.reduce((qd, item) => {let [k,v] = item.split`=`; v = v && decodeURIComponent(v); (qd[k] = qd[k] || []).push(v); return qd}, {})
    : {}
}

多行：

var qd = {};
if (location.search) location.search.substr(1).split("&").forEach(function(item) {
    var s = item.split("="),
        k = s[0],
        v = s[1] && decodeURIComponent(s[1]); //  null-coalescing / short-circuit
    //(k in qd) ? qd[k].push(v) : qd[k] = [v]
    (qd[k] = qd[k] || []).push(v) // null-coalescing / short-circuit
})

这是什么代码。。。“零合并”，短路评估ES6解构赋值、箭头函数、模板字符串####示例：

"?a=1&b=0&c=3&d&e&a=5&a=t%20e%20x%20t&e=http%3A%2F%2Fw3schools.com%2Fmy%20test.asp%3Fname%3Dståle%26car%3Dsaab"
> qd
a: ["1", "5", "t e x t"]
b: ["0"]
c: ["3"]
d: [undefined]
e: [undefined, "http://w3schools.com/my test.asp?name=ståle&car=saab"]

> qd.a[1]    // "5"
> qd["a"][1] // "5"

---

阅读更多。。。关于Vanilla JavaScript解决方案。

要访问URL的不同部分，请使用位置。（搜索|哈希）

最简单（虚拟）解决方案

var queryDict = {};
location.search.substr(1).split("&").forEach(function(item) {queryDict[item.split("=")[0]] = item.split("=")[1]})

正确处理空钥匙。使用找到的最后一个值覆盖多键。

"?a=1&b=0&c=3&d&e&a=5"
> queryDict
a: "5"
b: "0"
c: "3"
d: undefined
e: undefined

多值键

简单的密钥检查（字典中的项目）？dict.item.push（val）：dict.item=[val]

var qd = {};
location.search.substr(1).split("&").forEach(function(item) {(item.split("=")[0] in qd) ? qd[item.split("=")[0]].push(item.split("=")[1]) : qd[item.split("=")[0]] = [item.split("=")[1]]})

现在返回数组。按qd.key[index]或qd[key][index]访问值

> qd
a: ["1", "5"]
b: ["0"]
c: ["3"]
d: [undefined]
e: [undefined]

编码字符？

对第二次或两次拆分使用decodeURIComponent（）。

var qd = {};
location.search.substr(1).split("&").forEach(function(item) {var k = item.split("=")[0], v = decodeURIComponent(item.split("=")[1]); (k in qd) ? qd[k].push(v) : qd[k] = [v]})

####示例：

"?a=1&b=0&c=3&d&e&a=5&a=t%20e%20x%20t&e=http%3A%2F%2Fw3schools.com%2Fmy%20test.asp%3Fname%3Dståle%26car%3Dsaab"
> qd
a: ["1", "5", "t e x t"]
b: ["0"]
c: ["3"]
d: ["undefined"]  // decodeURIComponent(undefined) returns "undefined" !!!*
e: ["undefined", "http://w3schools.com/my test.asp?name=ståle&car=saab"]

---

# From comments **\*!!!** Please note, that `decodeURIComponent(undefined)` returns string `"undefined"`. The solution lies in a simple usage of [`&&`][5], which ensures that `decodeURIComponent()` is not called on undefined values. _(See the "complete solution" at the top.)_

v = v && decodeURIComponent(v);

If the querystring is empty (`location.search == ""`), the result is somewhat misleading `qd == {"": undefined}`. It is suggested to check the querystring before launching the parsing function likeso:

if (location.search) location.search.substr(1).split("&").forEach(...)