是否有一种通过jQuery(或不使用)检索查询字符串值的无插件方法?

如果是,怎么办?如果没有,是否有插件可以这样做?


当前回答

这里有一个很好的小url实用程序,带有一些很酷的糖霜:

http://www.example.com/path/index.html?silly=willy#chucky=cheese

url();            // http://www.example.com/path/index.html?silly=willy#chucky=cheese
url('domain');    // example.com
url('1');         // path
url('-1');        // index.html
url('?');         // silly=willy
url('?silly');    // willy
url('?poo');      // (an empty string)
url('#');         // chucky=cheese
url('#chucky');   // cheese
url('#poo');      // (an empty string)

查看更多示例并在此处下载:https://github.com/websanova/js-url#url

其他回答

如果您使用Browserify,则可以使用Node.js中的url模块:

var url = require('url');

url.parse('http://example.com/?bob=123', true).query;

// returns { "bob": "123" }

进一步阅读:URL Node.js v0.12.2手册和文档

编辑:您可以使用URL界面,它在几乎所有新浏览器中都被广泛采用,如果代码将在旧浏览器上运行,您可以使用像这样的polyfill。下面是一个关于如何使用URL接口获取查询参数(也称为搜索参数)的代码示例

const url = new URL('http://example.com/?bob=123');
url.searchParams.get('bob'); 

您也可以使用URLSearchParams进行搜索,下面是MDN的一个使用URLSearchParams进行搜索的示例:

var paramsString = "q=URLUtils.searchParams&topic=api";
var searchParams = new URLSearchParams(paramsString);

//Iterate the search parameters.
for (let p of searchParams) {
  console.log(p);
}

searchParams.has("topic") === true; // true
searchParams.get("topic") === "api"; // true
searchParams.getAll("topic"); // ["api"]
searchParams.get("foo") === null; // true
searchParams.append("topic", "webdev");
searchParams.toString(); // "q=URLUtils.searchParams&topic=api&topic=webdev"
searchParams.set("topic", "More webdev");
searchParams.toString(); // "q=URLUtils.searchParams&topic=More+webdev"
searchParams.delete("topic");
searchParams.toString(); // "q=URLUtils.searchParams"

使用纯JavaScript和正则表达式的简单解决方案:

alert(getQueryString("p2"));

function getQueryString (Param) {
    return decodeURI("http://www.example.com/?p1=p11&p2=p2222".replace(new RegExp("^(?:.*[&?]" + encodeURI(Param).replace(/[.+*]/g, "$&") + "(?:=([^&]*))?)?.*$", "i"), "$1"));
}

Js投标

Node.js的源代码中有一个健壮的实现https://github.com/joyent/node/blob/master/lib/querystring.js

TJ的qs也执行嵌套参数解析https://github.com/visionmedia/node-querystring

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;
};

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