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

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;
}

其他回答

我使用以下代码(JavaScript)获取通过URL传递的内容:

function getUrlVars() {
            var vars = {};
            var parts = window.location.href.replace(/[?&]+([^=&]+)=([^&]*)/gi, function(m,key,value) {
                vars[key] = value;
            });
            return vars;
        }

然后,要将值分配给变量,只需指定要获取的参数,例如,如果URL是example.com/?I=1&p=2&f=3

您可以执行此操作以获取值:

var getI = getUrlVars()["I"];
var getP = getUrlVars()["p"];
var getF = getUrlVars()["f"];

则值将为:

getI = 1, getP = 2 and getF = 3
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

我们刚刚发布了arg.js,这是一个旨在一劳永逸地解决这个问题的项目。传统上很难,但现在你可以做到:

var name = Arg.get("name");

或者得到全部:

var params = Arg.all();

如果你在乎两者之间的区别?query=true和#hash=true,则可以使用Arg.query()和Arg.hash()方法。

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;
};