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"]
}

/** * Parses and builds Object of URL query string. * @param {string} query The URL query string. * @return {!Object<string, string>} */ function parseQueryString(query) { if (!query) { return {}; } return (/^[?#]/.test(query) ? query.slice(1) : query) .split('&') .reduce((params, param) => { const item = param.split('='); const key = decodeURIComponent(item[0] || ''); const value = decodeURIComponent(item[1] || ''); if (key) { params[key] = value; } return params; }, {}); } console.log(parseQueryString('?v=MFa9pvnVe0w&ku=user&from=89&aw=1')) see log

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

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]);
}

2022 ES6/7/8，进近

从ES6开始，Javascript提供了几个构造来为这个问题创建一个性能解决方案。

这包括使用URLSearchParams和迭代器

let params = new URLSearchParams('abc=foo&def=%5Basf%5D&xyz=5');
params.get("abc"); // "foo"

如果你的用例需要你实际将其转换为对象，你可以实现以下函数:

function paramsToObject(entries) {
  const result = {}
  for(const [key, value] of entries) { // each 'entry' is a [key, value] tupple
    result[key] = value;
  }
  return result;
}

基本的演示

const urlParams = new URLSearchParams('abc=foo&def=%5Basf%5D&xyz=5');
const entries = urlParams.entries(); //returns an iterator of decoded [key,value] tuples
const params = paramsToObject(entries); //{abc:"foo",def:"[asf]",xyz:"5"}

使用Object.fromEntries和spread

我们可以使用Object.fromEntries，用Object.fromEntries(entries)替换paramsToObject。

对象的列表名称-值对是要遍历的值对 键是名称，值是值。

由于URLParams返回一个可迭代对象，使用展开操作符而不是调用.entries也将根据其规范生成条目:

const urlParams = new URLSearchParams('abc=foo&def=%5Basf%5D&xyz=5');
const params = Object.fromEntries(urlParams); // {abc: "foo", def: "[asf]", xyz: "5"}

注意:根据URLSearchParams规范，所有值都是自动字符串

多个相同的键

正如@siipe指出的，包含多个相同键值的字符串将被强制转换为最后一个可用值:foo=first_value&foo=second_value本质上将变成:{foo: "second_value"}。

根据这个答案:https://stackoverflow.com/a/1746566/1194694没有规范来决定用它做什么，每个框架可以有不同的行为。

一个常见的用例是将两个相同的值连接到一个数组中，使输出对象变成:

{foo: ["first_value", "second_value"]}

这可以通过以下代码实现:

const groupParamsByKey = (params) => [...params.entries()].reduce((acc, tuple) => {
 // getting the key and value from each tuple
 const [key, val] = tuple;
 if(acc.hasOwnProperty(key)) {
    // if the current key is already an array, we'll add the value to it
    if(Array.isArray(acc[key])) {
      acc[key] = [...acc[key], val]
    } else {
      // if it's not an array, but contains a value, we'll convert it into an array
      // and add the current value to it
      acc[key] = [acc[key], val];
    }
 } else {
  // plain assignment if no special case is present
  acc[key] = val;
 }

return acc;
}, {});

const params = new URLSearchParams('abc=foo&def=%5Basf%5D&xyz=5&def=dude');
const output = groupParamsByKey(params) // {abc: "foo", def: ["[asf]", "dude"], xyz: 5}

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; }