在Mike Causer回答的基础上，我创建了这个函数，它考虑了具有相同键的多个参数(foo=bar&foo=baz)和逗号分隔的参数(foo=bar,baz,bin)。它还允许您搜索某个查询键。

function getQueryParams(queryKey) {
    var queryString = window.location.search;
    var query = {};
    var pairs = (queryString[0] === '?' ? queryString.substr(1) : queryString).split('&');
    for (var i = 0; i < pairs.length; i++) {
        var pair = pairs[i].split('=');
        var key = decodeURIComponent(pair[0]);
        var value = decodeURIComponent(pair[1] || '');
        // Se possui uma vírgula no valor, converter em um array
        value = (value.indexOf(',') === -1 ? value : value.split(','));

        // Se a key já existe, tratar ela como um array
        if (query[key]) {
            if (query[key].constructor === Array) {
                // Array.concat() faz merge se o valor inserido for um array
                query[key] = query[key].concat(value);
            } else {
                // Se não for um array, criar um array contendo o valor anterior e o novo valor
                query[key] = [query[key], value];
            }
        } else {
            query[key] = value;
        }
    }

    if (typeof queryKey === 'undefined') {
        return query;
    } else {
        return query[queryKey];
    }
}

示例输入: foo.html吗?博兹,foo = bar&foo = baz&foo =鹿角的第二叉,buz&bar = 1, 2, 3

示例输出

{
    foo: ["bar","baz","bez","boz","buz"],
    bar: ["1","2","3"]
}

下面是我用的一个例子:

var params = {};
window.location.search.substring(1).split('&').forEach(function(pair) {
  pair = pair.split('=');
  if (pair[1] !== undefined) {
    var key = decodeURIComponent(pair[0]),
        val = decodeURIComponent(pair[1]),
        val = val ? val.replace(/\++/g,' ').trim() : '';

    if (key.length === 0) {
      return;
    }
    if (params[key] === undefined) {
      params[key] = val;
    }
    else {
      if ("function" !== typeof params[key].push) {
        params[key] = [params[key]];
      }
      params[key].push(val);
    }
  }
});
console.log(params);

基本用法。 ? = aa&b = bb 对象{a: "aa"， b: "bb"}

重复参数，例如。 ? = aa&b = bb&c = cc&c =土豆 对象{a: "aa"， b: "bb"， c: ["cc"，"potato"]}

钥匙不见了。 ? = aa&b = bb = cc 对象{a: "aa"， b: "bb"}

缺少值，例如。 = aa&b = bb&c ? 对象{a: "aa"， b: "bb"}

上述JSON/regex解决方案在这个古怪的url上抛出了一个语法错误: ? = aa&b = bb&c = & = dd&e 对象{a: "aa"， b: "bb"， c: ""}

我发现$. string .deparam是最完整的预构建解决方案(可以做嵌套对象等)。查看文档。

到目前为止，我发现的建议解决方案并没有涵盖更复杂的场景。

我需要像这样转换查询字符串

https://random.url.com?Target=Offer&Method=findAll&filters%5Bhas_goals_enabled%5D%5BTRUE%5D=1&filters%5Bstatus%5D=active&fields%5B%5D=id&fields%5B%5D=name&fields%5B%5D=default_goal_name

变成一个像这样的物体:

{
    "Target": "Offer",
    "Method": "findAll",
    "fields": [
        "id",
        "name",
        "default_goal_name"
    ],
    "filters": {
        "has_goals_enabled": {
            "TRUE": "1"
        },
        "status": "active"
    }
}

OR:

https://random.url.com?Target=Report&Method=getStats&fields%5B%5D=Offer.name&fields%5B%5D=Advertiser.company&fields%5B%5D=Stat.clicks&fields%5B%5D=Stat.conversions&fields%5B%5D=Stat.cpa&fields%5B%5D=Stat.payout&fields%5B%5D=Stat.date&fields%5B%5D=Stat.offer_id&fields%5B%5D=Affiliate.company&groups%5B%5D=Stat.offer_id&groups%5B%5D=Stat.date&filters%5BStat.affiliate_id%5D%5Bconditional%5D=EQUAL_TO&filters%5BStat.affiliate_id%5D%5Bvalues%5D=1831&limit=9999

成:

{
    "Target": "Report",
    "Method": "getStats",
    "fields": [
        "Offer.name",
        "Advertiser.company",
        "Stat.clicks",
        "Stat.conversions",
        "Stat.cpa",
        "Stat.payout",
        "Stat.date",
        "Stat.offer_id",
        "Affiliate.company"
    ],
    "groups": [
        "Stat.offer_id",
        "Stat.date"
    ],
    "limit": "9999",
    "filters": {
        "Stat.affiliate_id": {
            "conditional": "EQUAL_TO",
            "values": "1831"
        }
    }
}

我将多个解决方案编译并调整为一个实际有效的解决方案:

代码:

var getParamsAsObject = function (query) {

    query = query.substring(query.indexOf('?') + 1);

    var re = /([^&=]+)=?([^&]*)/g;
    var decodeRE = /\+/g;

    var decode = function (str) {
        return decodeURIComponent(str.replace(decodeRE, " "));
    };

    var params = {}, e;
    while (e = re.exec(query)) {
        var k = decode(e[1]), v = decode(e[2]);
        if (k.substring(k.length - 2) === '[]') {
            k = k.substring(0, k.length - 2);
            (params[k] || (params[k] = [])).push(v);
        }
        else params[k] = v;
    }

    var assign = function (obj, keyPath, value) {
        var lastKeyIndex = keyPath.length - 1;
        for (var i = 0; i < lastKeyIndex; ++i) {
            var key = keyPath[i];
            if (!(key in obj))
                obj[key] = {}
            obj = obj[key];
        }
        obj[keyPath[lastKeyIndex]] = value;
    }

    for (var prop in params) {
        var structure = prop.split('[');
        if (structure.length > 1) {
            var levels = [];
            structure.forEach(function (item, i) {
                var key = item.replace(/[?[\]\\ ]/g, '');
                levels.push(key);
            });
            assign(params, levels, params[prop]);
            delete(params[prop]);
        }
    }
    return params;
};

一个班轮。干净简单。

const params = Object.fromEntries(new URLSearchParams(location.search));

对于您的具体情况，它将是:

const str = 'abc=foo&def=%5Basf%5D&xyz=5'; const params = Object.fromEntries(new URLSearchParams(str)); console.log (params);

在&上拆分以获得名称/值对，然后在=上拆分每对。这里有一个例子:

var str = "abc=foo&def=%5Basf%5D&xy%5Bz=5"
var obj = str.split("&").reduce(function(prev, curr, i, arr) {
    var p = curr.split("=");
    prev[decodeURIComponent(p[0])] = decodeURIComponent(p[1]);
    return prev;
}, {});

另一种方法，使用正则表达式:

var obj = {}; 
str.replace(/([^=&]+)=([^&]*)/g, function(m, key, value) {
    obj[decodeURIComponent(key)] = decodeURIComponent(value);
}); 

本文改编自约翰·瑞西格的《搜索和不替换》。