下面是硅制品方法的一个更简化的版本。

下面的函数可以从USVString或Location解析查询字符串。

/** * Returns a plain object representation of a URLSearchParams object. * @param {USVString} search - A URL querystring * @return {Object} a key-value pair object from a URL querystring */ const parseSearch = (search) => [...new URLSearchParams(search).entries()] .reduce((acc, [key, val]) => ({ ...acc, // eslint-disable-next-line no-nested-ternary [key]: Object.prototype.hasOwnProperty.call(acc, key) ? Array.isArray(acc[key]) ? [...acc[key], val] : [acc[key], val] : val }), {}); /** * Returns a plain object representation of a URLSearchParams object. * @param {Location} location - Either a document or window location, or React useLocation() * @return {Object} a key-value pair object from a URL querystring */ const parseLocationSearch = (location) => parseSearch(location.search); console.log(parseSearch('?foo=bar&x=y&ids=%5B1%2C2%2C3%5D&ids=%5B4%2C5%2C6%5D')); .as-console-wrapper { top: 0; max-height: 100% !important; }

下面是上面代码的一行代码(125字节):

f是parsearchch

f=s=>[...new URLSearchParams(s).entries()].reduce((a,[k,v])=>({...a,[k]:a[k]?Array.isArray(a[k])?[...a[k],v]:[a[k],v]:v}),{})

Edit

下面是一个序列化和更新的方法:

const parseSearch = (search) => [...new URLSearchParams(search).entries()] .reduce((acc, [key, val]) => ({ ...acc, // eslint-disable-next-line no-nested-ternary [key]: Object.prototype.hasOwnProperty.call(acc, key) ? Array.isArray(acc[key]) ? [...acc[key], val] : [acc[key], val] : val }), {}); const toQueryString = (params) => `?${Object.entries(params) .flatMap(([key, values]) => Array.isArray(values) ? values.map(value => [key, value]) : [[key, values]]) .map(pair => pair.map(val => encodeURIComponent(val)).join('=')) .join('&')}`; const updateQueryString = (search, update) => (parsed => toQueryString(update instanceof Function ? update(parsed) : { ...parsed, ...update })) (parseSearch(search)); const queryString = '?foo=bar&x=y&ids=%5B1%2C2%2C3%5D&ids=%5B4%2C5%2C6%5D'; const parsedQuery = parseSearch(queryString); console.log(parsedQuery); console.log(toQueryString(parsedQuery) === queryString); const updatedQuerySimple = updateQueryString(queryString, { foo: 'baz', x: 'z', }); console.log(updatedQuerySimple); console.log(parseSearch(updatedQuerySimple)); const updatedQuery = updateQueryString(updatedQuerySimple, parsed => ({ ...parsed, ids: [ ...parsed.ids, JSON.stringify([7,8,9]) ] })); console.log(updatedQuery); console.log(parseSearch(updatedQuery)); .as-console-wrapper { top: 0; max-height: 100% !important; }

这是一个简单的版本，显然你需要添加一些错误检查:

var obj = {};
var pairs = queryString.split('&');
for(i in pairs){
    var split = pairs[i].split('=');
    obj[decodeURIComponent(split[0])] = decodeURIComponent(split[1]);
}

使用URLSearchParams JavaScript Web API非常简单，

var paramsString = "abc=foo&def=%5Basf%5D&xyz=5"; //returns an iterator object var searchParams = new URLSearchParams(paramsString); //Usage for (let p of searchParams) { console.log(p); } //Get the query strings console.log(searchParams.toString()); //You can also pass in objects var paramsObject = {abc:"forum",def:"%5Basf%5D",xyz:"5"} //returns an iterator object var searchParams = new URLSearchParams(paramsObject); //Usage for (let p of searchParams) { console.log(p); } //Get the query strings console.log(searchParams.toString());

# #的有用链接

URLSearchParams - Web api | MDN 简单的URL操作与URLSearchParams | Web |谷歌开发者

注意:IE不支持

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,"\":\"")) + '"}')

我也遇到了同样的问题，尝试了这里的解决方案，但没有一个真正有效，因为我在URL参数中有数组，像这样:

?param[]=5&param[]=8&othr_param=abc&param[]=string

所以我最终写了我自己的JS函数，它使一个数组的参数在URI:

/**
 * Creates an object from URL encoded data
 */
var createObjFromURI = function() {
    var uri = decodeURI(location.search.substr(1));
    var chunks = uri.split('&');
    var params = Object();

    for (var i=0; i < chunks.length ; i++) {
        var chunk = chunks[i].split('=');
        if(chunk[0].search("\\[\\]") !== -1) {
            if( typeof params[chunk[0]] === 'undefined' ) {
                params[chunk[0]] = [chunk[1]];

            } else {
                params[chunk[0]].push(chunk[1]);
            }


        } else {
            params[chunk[0]] = chunk[1];
        }
    }

    return 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》。