ES6一行(如果我们可以这样称呼它，看到长行)

[…新URLSearchParams (location.search) .entries())。Reduce ((prev， [key,val]) => {prev[key] = val;返回prev}， {})

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;
};

据我所知，没有原生的解决方案。Dojo有一个内置的反序列化方法(如果您碰巧使用了该框架)。

否则，你可以自己简单地实现它:

function unserialize(str) {
  str = decodeURIComponent(str);
  var chunks = str.split('&'),
      obj = {};
  for(var c=0; c < chunks.length; c++) {
    var split = chunks[c].split('=', 2);
    obj[split[0]] = split[1];
  }
  return obj;
}

编辑:添加decodeURIComponent()

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》。

在2021年…请认为这是过时的。

Edit

这个编辑改进并解释了基于评论的答案。

var search = location.search.substring(1);
JSON.parse('{"' + decodeURI(search).replace(/"/g, '\\"').replace(/&/g, '","').replace(/=/g,'":"') + '"}')

例子

分五个步骤解析abc=foo&def=%5Basf%5D&xyz=5:

decodeURI: abc = foo&def = xyz (asf) = 5 转义引号:相同，因为没有引号 替换&:abc=foo"，"def=[asf]"，"xyz=5 " 5 . Replace =: abc":"foo"，"def":"[asf]"，"xyz": Suround卷曲和引用:{“abc”:“foo”、“def”:“(asf)”,“xyz”:“5”}

这是合法的JSON。

改进的解决方案允许搜索字符串中有更多字符。它使用了一个恢复函数来进行URI解码:

var search = location.search.substring(1);
JSON.parse('{"' + search.replace(/&/g, '","').replace(/=/g,'":"') + '"}', function(key, value) { return key===""?value:decodeURIComponent(value) })

例子

search = "abc=foo&def=%5Basf%5D&xyz=5&foo=b%3Dar";

给了

Object {abc: "foo", def: "[asf]", xyz: "5", foo: "b=ar"}

原来的答案

一行程序:

JSON.parse('{"' + decodeURI("abc=foo&def=%5Basf%5D&xyz=5".replace(/&/g, "\",\"").replace(/=/g,"\":\"")) + '"}')

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。