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

ES2015(ES6)

getQueryStringParams = query => {
    return query
        ? (/^[?#]/.test(query) ? query.slice(1) : query)
            .split('&')
            .reduce((params, param) => {
                    let [key, value] = param.split('=');
                    params[key] = value ? decodeURIComponent(value.replace(/\+/g, ' ')) : '';
                    return params;
                }, {}
            )
        : {}
};

没有jQuery

var qs = (function(a) {
    if (a == "") return {};
    var b = {};
    for (var i = 0; i < a.length; ++i)
    {
        var p=a[i].split('=', 2);
        if (p.length == 1)
            b[p[0]] = "";
        else
            b[p[0]] = decodeURIComponent(p[1].replace(/\+/g, " "));
    }
    return b;
})(window.location.search.substr(1).split('&'));

URL如下?topic=123&name=query+string,将返回以下内容:

qs["topic"];    // 123
qs["name"];     // query string
qs["nothere"];  // undefined (object)

Google方法

撕扯谷歌的代码,我找到了他们使用的方法:getUrlParameters

function (b) {
    var c = typeof b === "undefined";
    if (a !== h && c) return a;
    for (var d = {}, b = b || k[B][vb], e = b[p]("?"), f = b[p]("#"), b = (f === -1 ? b[Ya](e + 1) : [b[Ya](e + 1, f - e - 1), "&", b[Ya](f + 1)][K](""))[z]("&"), e = i.dd ? ia : unescape, f = 0, g = b[w]; f < g; ++f) {
        var l = b[f][p]("=");
        if (l !== -1) {
            var q = b[f][I](0, l),
                l = b[f][I](l + 1),
                l = l[Ca](/\+/g, " ");
            try {
                d[q] = e(l)
            } catch (A) {}
        }
    }
    c && (a = d);
    return d
}

这是模糊的,但可以理解。它无法工作,因为某些变量未定义。

他们开始在url上查找参数?并且还从散列#中。然后,对于每个参数,它们以等号b[f][p](“=”)分割(看起来像indexOf,它们使用字符的位置来获取键/值)。拆分后,他们检查参数是否有值,如果有值,则存储d的值,否则继续。

最后返回对象d,处理转义和+符号。这个对象和我的一样,它有相同的行为。


我的方法作为jQuery插件

(function($) {
    $.QueryString = (function(paramsArray) {
        let params = {};

        for (let i = 0; i < paramsArray.length; ++i)
        {
            let param = paramsArray[i]
                .split('=', 2);
            
            if (param.length !== 2)
                continue;
            
            params[param[0]] = decodeURIComponent(param[1].replace(/\+/g, " "));
        }
            
        return params;
    })(window.location.search.substr(1).split('&'))
})(jQuery);

用法

//Get a param
$.QueryString.param
//-or-
$.QueryString["param"]
//This outputs something like...
//"val"

//Get all params as object
$.QueryString
//This outputs something like...
//Object { param: "val", param2: "val" }

//Set a param (only in the $.QueryString object, doesn't affect the browser's querystring)
$.QueryString.param = "newvalue"
//This doesn't output anything, it just updates the $.QueryString object

//Convert object into string suitable for url a querystring (Requires jQuery)
$.param($.QueryString)
//This outputs something like...
//"param=newvalue&param2=val"

//Update the url/querystring in the browser's location bar with the $.QueryString object
history.replaceState({}, '', "?" + $.param($.QueryString));
//-or-
history.pushState({}, '', "?" + $.param($.QueryString));

性能测试(针对正则表达式方法的拆分方法)(jsPerf)

准备代码:方法声明

拆分测试代码

var qs = window.GetQueryString(query);

var search = qs["q"];
var value = qs["value"];
var undef = qs["undefinedstring"];

Regex测试代码

var search = window.getParameterByName("q");
var value = window.getParameterByName("value");
var undef = window.getParameterByName("undefinedstring");

在Windows Server 2008 R2/7 x64上的Firefox 4.0 x86中测试

拆分方法:最快144780±2.17%Regex方法:13891±0.85%|90%慢

其他回答

这是我在GitHub上的查询字符串解析代码版本。

它的前缀是jquery.*,但解析函数本身不使用jquery。它非常快,但仍然可以进行一些简单的性能优化。

它还支持URL中的列表和哈希表编码,例如:

arr[]=10&arr[]=20&arr[]=100

or

hash[key1]=hello&hash[key2]=moto&a=How%20are%20you

jQuery.toQueryParams = function(str, separator) {
    separator = separator || '&'
    var obj = {}
    if (str.length == 0)
        return obj
    var c = str.substr(0,1)
    var s = c=='?' || c=='#'  ? str.substr(1) : str; 

    var a = s.split(separator)
    for (var i=0; i<a.length; i++) {
        var p = a[i].indexOf('=')
        if (p < 0) {
            obj[a[i]] = ''
            continue
        }
        var k = decodeURIComponent(a[i].substr(0,p)),
            v = decodeURIComponent(a[i].substr(p+1))

        var bps = k.indexOf('[')
        if (bps < 0) {
            obj[k] = v
            continue;
        } 

        var bpe = k.substr(bps+1).indexOf(']')
        if (bpe < 0) {
            obj[k] = v
            continue;
        }

        var bpv = k.substr(bps+1, bps+bpe-1)
        var k = k.substr(0,bps)
        if (bpv.length <= 0) {
            if (typeof(obj[k]) != 'object') obj[k] = []
            obj[k].push(v)
        } else {
            if (typeof(obj[k]) != 'object') obj[k] = {}
            obj[k][bpv] = v
        }
    }
    return obj;

}

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;
};

可靠地做这件事比一开始想象的要复杂得多。

其他答案中使用的location.search很脆弱,应该避免使用-例如,如果有人搞砸了,并在?查询字符串。在我看来,URL在浏览器中自动转义的方式有很多种,这使得decodeURIComponent非常强制性。许多查询字符串是由用户输入生成的,这意味着对URL内容的假设非常糟糕。包括非常基本的东西,比如每个键都是唯一的,甚至有一个值。

为了解决这个问题,这里提供了一个可配置的API,并提供了健康的防御性编程。请注意,如果您愿意对某些变量进行硬编码,或者如果输入不能包含hasOwnProperty等,则可以将其大小减半。

版本1:返回包含每个参数的名称和值的数据对象。它有效地消除了重复,并始终尊重从左到右找到的第一个。

function getQueryData(url, paramKey, pairKey, missingValue, decode) {

    var query, queryStart, fragStart, pairKeyStart, i, len, name, value, result;

    if (!url || typeof url !== 'string') {
        url = location.href; // more robust than location.search, which is flaky
    }
    if (!paramKey || typeof paramKey !== 'string') {
        paramKey = '&';
    }
    if (!pairKey || typeof pairKey !== 'string') {
        pairKey = '=';
    }
    // when you do not explicitly tell the API...
    if (arguments.length < 5) {
        // it will unescape parameter keys and values by default...
        decode = true;
    }

    queryStart = url.indexOf('?');
    if (queryStart >= 0) {
        // grab everything after the very first ? question mark...
        query = url.substring(queryStart + 1);
    } else {
        // assume the input is already parameter data...
        query = url;
    }
    // remove fragment identifiers...
    fragStart = query.indexOf('#');
    if (fragStart >= 0) {
        // remove everything after the first # hash mark...
        query = query.substring(0, fragStart);
    }
    // make sure at this point we have enough material to do something useful...
    if (query.indexOf(paramKey) >= 0 || query.indexOf(pairKey) >= 0) {
        // we no longer need the whole query, so get the parameters...
        query = query.split(paramKey);
        result = {};
        // loop through the parameters...
        for (i = 0, len = query.length; i < len; i = i + 1) {
            pairKeyStart = query[i].indexOf(pairKey);
            if (pairKeyStart >= 0) {
                name = query[i].substring(0, pairKeyStart);
            } else {
                name = query[i];
            }
            // only continue for non-empty names that we have not seen before...
            if (name && !Object.prototype.hasOwnProperty.call(result, name)) {
                if (decode) {
                    // unescape characters with special meaning like ? and #
                    name = decodeURIComponent(name);
                }
                if (pairKeyStart >= 0) {
                    value = query[i].substring(pairKeyStart + 1);
                    if (value) {
                        if (decode) {
                            value = decodeURIComponent(value);
                        }
                    } else {
                        value = missingValue;
                    }
                } else {
                    value = missingValue;
                }
                result[name] = value;
            }
        }
        return result;
    }
}

版本2:返回一个具有两个相同长度数组的数据映射对象,一个用于名称,另一个用于值,每个参数都有一个索引。此格式支持重复名称,并故意不消除重复名称,因为这可能就是您希望使用此格式的原因。

function getQueryData(url, paramKey, pairKey, missingValue, decode) {

    var query, queryStart, fragStart, pairKeyStart, i, len, name, value, result;

    if (!url || typeof url !== 'string') {
          url = location.href; // more robust than location.search, which is flaky
    }
        if (!paramKey || typeof paramKey !== 'string') {
            paramKey = '&';
        }
        if (!pairKey || typeof pairKey !== 'string') {
            pairKey = '=';
        }
        // when you do not explicitly tell the API...
        if (arguments.length < 5) {
            // it will unescape parameter keys and values by default...
            decode = true;
        }

        queryStart = url.indexOf('?');
        if (queryStart >= 0) {
            // grab everything after the very first ? question mark...
            query = url.substring(queryStart + 1);
        } else {
            // assume the input is already parameter data...
            query = url;
        }
        // remove fragment identifiers...
        fragStart = query.indexOf('#');
        if (fragStart >= 0) {
            // remove everything after the first # hash mark...
            query = query.substring(0, fragStart);
        }
        // make sure at this point we have enough material to do something useful...
        if (query.indexOf(paramKey) >= 0 || query.indexOf(pairKey) >= 0) {
            // we no longer need the whole query, so get the parameters...
            query = query.split(paramKey);
            result = {
                names: [],
                values: []
            };
            // loop through the parameters...
            for (i = 0, len = query.length; i < len; i = i + 1) {
                pairKeyStart = query[i].indexOf(pairKey);
                if (pairKeyStart >= 0) {
                    name = query[i].substring(0, pairKeyStart);
                } else {
                    name = query[i];
                }
                // only continue for non-empty names...
                if (name) {
                    if (decode) {
                        // unescape characters with special meaning like ? and #
                        name = decodeURIComponent(name);
                    }
                    if (pairKeyStart >= 0) {
                        value = query[i].substring(pairKeyStart + 1);
                        if (value) {
                            if (decode) {
                                value = decodeURIComponent(value);
                            }
                        } else {
                            value = missingValue;
                        }
                    } else {
                        value = missingValue;
                    }
                    result.names.push(name);
                    result.values.push(value);
                }
           }
           return result;
       }
   }

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;
};

关于这个问题的顶级答案的问题是,不支持在#后面放置参数,但有时也需要获得这个值。

我修改了答案,让它解析带有哈希符号的完整查询字符串:

var getQueryStringData = function(name) {
    var result = null;
    var regexS = "[\\?&#]" + name + "=([^&#]*)";
    var regex = new RegExp(regexS);
    var results = regex.exec('?' + window.location.href.split('?')[1]);
    if (results != null) {
        result = decodeURIComponent(results[1].replace(/\+/g, " "));
    }
    return result;
};