我有一个这样的字符串:
abc=foo&def=%5Basf%5D&xyz=5
如何将其转换为这样的JavaScript对象?
{
abc: 'foo',
def: '[asf]',
xyz: 5
}
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]);
}
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》。
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,"\":\"")) + '"}')
首先你需要定义什么是get变量:
function getVar()
{
this.length = 0;
this.keys = [];
this.push = function(key, value)
{
if(key=="") key = this.length++;
this[key] = value;
this.keys.push(key);
return this[key];
}
}
而不是直接读:
function urlElement()
{
var thisPrototype = window.location;
for(var prototypeI in thisPrototype) this[prototypeI] = thisPrototype[prototypeI];
this.Variables = new getVar();
if(!this.search) return this;
var variables = this.search.replace(/\?/g,'').split('&');
for(var varI=0; varI<variables.length; varI++)
{
var nameval = variables[varI].split('=');
var name = nameval[0].replace(/\]/g,'').split('[');
var pVariable = this.Variables;
for(var nameI=0;nameI<name.length;nameI++)
{
if(name.length-1==nameI) pVariable.push(name[nameI],nameval[1]);
else var pVariable = (typeof pVariable[name[nameI]] != 'object')? pVariable.push(name[nameI],new getVar()) : pVariable[name[nameI]];
}
}
}
并使用like:
var mlocation = new urlElement();
mlocation = mlocation.Variables;
for(var key=0;key<mlocation.keys.length;key++)
{
console.log(key);
console.log(mlocation[mlocation.keys[key]];
}
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¶m[]=8&othr_param=abc¶m[]=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;
}
这似乎是最好的解决方案，因为它考虑了同名的多个参数。
function paramsToJSON(str) {
var pairs = str.split('&');
var result = {};
pairs.forEach(function(pair) {
pair = pair.split('=');
var name = pair[0]
var value = pair[1]
if( name.length )
if (result[name] !== undefined) {
if (!result[name].push) {
result[name] = [result[name]];
}
result[name].push(value || '');
} else {
result[name] = value || '';
}
});
return( result );
}
<a href="index.html?x=1&x=2&x=3&y=blah">something</a>
paramsToJSON("x=1&x=2&x=3&y=blah");
console yields => {x: Array[3], y: "blah"} where x is an array as is proper JSON
后来我决定把它转换成一个jQuery插件…
$.fn.serializeURLParams = function() {
var result = {};
if( !this.is("a") || this.attr("href").indexOf("?") == -1 )
return( result );
var pairs = this.attr("href").split("?")[1].split('&');
pairs.forEach(function(pair) {
pair = pair.split('=');
var name = decodeURI(pair[0])
var value = decodeURI(pair[1])
if( name.length )
if (result[name] !== undefined) {
if (!result[name].push) {
result[name] = [result[name]];
}
result[name].push(value || '');
} else {
result[name] = value || '';
}
});
return( result )
}
<a href="index.html?x=1&x=2&x=3&y=blah">something</a>
$("a").serializeURLParams();
console yields => {x: Array[3], y: "blah"} where x is an array as is proper JSON
现在，第一个将只接受参数，但jQuery插件将接受整个url并返回序列化的参数。
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: ""}
有一个名为YouAreI.js的轻量级库，它经过测试，使这个非常简单。
YouAreI = require('YouAreI')
uri = new YouAreI('http://user:pass@www.example.com:3000/a/b/c?d=dad&e=1&f=12.3#fragment');
uri.query_get() => { d: 'dad', e: '1', f: '12.3' }
另一种基于URLSearchParams最新标准的解决方案(https://developer.mozilla.org/en-US/docs/Web/API/URLSearchParams)
function getQueryParamsObject() {
const searchParams = new URLSearchParams(location.search.slice(1));
return searchParams
? _.fromPairs(Array.from(searchParams.entries()))
: {};
}
请注意，这个解决方案是利用
Array.from (https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/from)
和lodash的_. fropairs (https://lodash.com/docs#fromPairs)，以便简单。
因为您可以访问searchParams.entries()迭代器，所以创建一个更兼容的解决方案应该很容易。
我还需要在URL的查询部分处理+ (decodeURIComponent没有)，所以我改编了Wolfgang的代码成为:
var search = location.search.substring(1);
search = search?JSON.parse('{"' + search.replace(/\+/g, ' ').replace(/&/g, '","').replace(/=/g,'":"') + '"}',
function(key, value) { return key===""?value:decodeURIComponent(value)}):{};
在我的例子中，我使用jQuery来获得URL准备好的表单参数，然后这个技巧来构建一个对象，然后我可以轻松地更新对象上的参数并重新构建查询URL，例如:
var objForm = JSON.parse('{"' + $myForm.serialize().replace(/\+/g, ' ').replace(/&/g, '","').replace(/=/g,'":"') + '"}',
function(key, value) { return key===""?value:decodeURIComponent(value)});
objForm.anyParam += stringToAddToTheParam;
var serializedForm = $.param(objForm);
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;
};
下面是我的快速而粗糙的版本，基本上它将以'&'分隔的URL参数拆分为数组元素，然后迭代该数组，将以'='分隔的键/值对添加到一个对象中。我使用decodeURIComponent()将编码字符转换为正常的字符串等效(因此%20变成空格，%26变成'&'等):
function deparam(paramStr) {
let paramArr = paramStr.split('&');
let paramObj = {};
paramArr.forEach(e=>{
let param = e.split('=');
paramObj[param[0]] = decodeURIComponent(param[1]);
});
return paramObj;
}
例子:
deparam('abc=foo&def=%5Basf%5D&xyz=5')
返回
{
abc: "foo"
def:"[asf]"
xyz :"5"
}
唯一的问题是xyz是一个字符串而不是一个数字(由于使用decodeURIComponent())，但除此之外，它不是一个坏的起点。
//under ES6
const getUrlParamAsObject = (url = window.location.href) => {
let searchParams = url.split('?')[1];
const result = {};
//in case the queryString is empty
if (searchParams!==undefined) {
const paramParts = searchParams.split('&');
for(let part of paramParts) {
let paramValuePair = part.split('=');
//exclude the case when the param has no value
if(paramValuePair.length===2) {
result[paramValuePair[0]] = decodeURIComponent(paramValuePair[1]);
}
}
}
return result;
}
使用phpjs
function parse_str(str, array) {
// discuss at: http://phpjs.org/functions/parse_str/
// original by: Cagri Ekin
// improved by: Michael White (http://getsprink.com)
// improved by: Jack
// improved by: Brett Zamir (http://brett-zamir.me)
// bugfixed by: Onno Marsman
// bugfixed by: Brett Zamir (http://brett-zamir.me)
// bugfixed by: stag019
// bugfixed by: Brett Zamir (http://brett-zamir.me)
// bugfixed by: MIO_KODUKI (http://mio-koduki.blogspot.com/)
// reimplemented by: stag019
// input by: Dreamer
// input by: Zaide (http://zaidesthings.com/)
// input by: David Pesta (http://davidpesta.com/)
// input by: jeicquest
// note: When no argument is specified, will put variables in global scope.
// note: When a particular argument has been passed, and the returned value is different parse_str of PHP. For example, a=b=c&d====c
// test: skip
// example 1: var arr = {};
// example 1: parse_str('first=foo&second=bar', arr);
// example 1: $result = arr
// returns 1: { first: 'foo', second: 'bar' }
// example 2: var arr = {};
// example 2: parse_str('str_a=Jack+and+Jill+didn%27t+see+the+well.', arr);
// example 2: $result = arr
// returns 2: { str_a: "Jack and Jill didn't see the well." }
// example 3: var abc = {3:'a'};
// example 3: parse_str('abc[a][b]["c"]=def&abc[q]=t+5');
// returns 3: {"3":"a","a":{"b":{"c":"def"}},"q":"t 5"}
var strArr = String(str)
.replace(/^&/, '')
.replace(/&$/, '')
.split('&'),
sal = strArr.length,
i, j, ct, p, lastObj, obj, lastIter, undef, chr, tmp, key, value,
postLeftBracketPos, keys, keysLen,
fixStr = function(str) {
return decodeURIComponent(str.replace(/\+/g, '%20'));
};
if (!array) {
array = this.window;
}
for (i = 0; i < sal; i++) {
tmp = strArr[i].split('=');
key = fixStr(tmp[0]);
value = (tmp.length < 2) ? '' : fixStr(tmp[1]);
while (key.charAt(0) === ' ') {
key = key.slice(1);
}
if (key.indexOf('\x00') > -1) {
key = key.slice(0, key.indexOf('\x00'));
}
if (key && key.charAt(0) !== '[') {
keys = [];
postLeftBracketPos = 0;
for (j = 0; j < key.length; j++) {
if (key.charAt(j) === '[' && !postLeftBracketPos) {
postLeftBracketPos = j + 1;
} else if (key.charAt(j) === ']') {
if (postLeftBracketPos) {
if (!keys.length) {
keys.push(key.slice(0, postLeftBracketPos - 1));
}
keys.push(key.substr(postLeftBracketPos, j - postLeftBracketPos));
postLeftBracketPos = 0;
if (key.charAt(j + 1) !== '[') {
break;
}
}
}
}
if (!keys.length) {
keys = [key];
}
for (j = 0; j < keys[0].length; j++) {
chr = keys[0].charAt(j);
if (chr === ' ' || chr === '.' || chr === '[') {
keys[0] = keys[0].substr(0, j) + '_' + keys[0].substr(j + 1);
}
if (chr === '[') {
break;
}
}
obj = array;
for (j = 0, keysLen = keys.length; j < keysLen; j++) {
key = keys[j].replace(/^['"]/, '')
.replace(/['"]$/, '');
lastIter = j !== keys.length - 1;
lastObj = obj;
if ((key !== '' && key !== ' ') || j === 0) {
if (obj[key] === undef) {
obj[key] = {};
}
obj = obj[key];
} else { // To insert new dimension
ct = -1;
for (p in obj) {
if (obj.hasOwnProperty(p)) {
if (+p > ct && p.match(/^\d+$/g)) {
ct = +p;
}
}
}
key = ct + 1;
}
}
lastObj[key] = value;
}
}
}
使用ES6, API和URLSearchParams。
function objectifyQueryString(url) {
let _url = new URL(url);
let _params = new URLSearchParams(_url.search);
let query = Array.from(_params.keys()).reduce((sum, value)=>{
return Object.assign({[value]: _params.get(value)}, sum);
}, {});
return query;
}
一个简明的解决方案:
location.search
.slice(1)
.split('&')
.map(p => p.split('='))
.reduce((obj, pair) => {
const [key, value] = pair.map(decodeURIComponent);
obj[key] = value;
return obj;
}, {});
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"]
}
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);
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
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}
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}， {})
如果你正在使用URI.js，你可以使用:
https://medialize.github.io/URI.js/docs.html#static-parseQuery
var result = URI.parseQuery("?foo=bar&hello=world&hello=mars&bam=&yup");
result === {
foo: "bar",
hello: ["world", "mars"],
bam: "",
yup: null
};
对于Node JS，你可以使用Node JS API querystring:
const querystring = require('querystring');
querystring.parse('abc=foo&def=%5Basf%5D&xyz=5&foo=b%3Dar');
// returns the object
文档:https://nodejs.org/api/querystring.html
2023年一行方法
一般情况下，你想要解析查询参数到一个对象:
Object.fromEntries(new URLSearchParams(location.search));
针对您的具体情况:
Object.fromEntries(new URLSearchParams('abc=foo&def=%5Basf%5D&xyz=5'));
console.log (decodeURI (' abc = foo&def = % 5巴斯夫% 5 d&xyz = 5 ') .split (' & ') .reduce((result, current) => { Const [key, value] = current.split('='); 结果[key] = value; 返回结果 }, {}))
如果您需要递归，您可以使用小巧的js-extension-ling库。
npm i js-extension-ling
const jsx = require("js-extension-ling");
console.log(jsx.queryStringToObject("a=1"));
console.log(jsx.queryStringToObject("a=1&a=3"));
console.log(jsx.queryStringToObject("a[]=1"));
console.log(jsx.queryStringToObject("a[]=1&a[]=pomme"));
console.log(jsx.queryStringToObject("a[0]=one&a[1]=five"));
console.log(jsx.queryStringToObject("http://blabla?foo=bar&number=1234"));
console.log(jsx.queryStringToObject("a[fruits][red][]=strawberry"));
console.log(jsx.queryStringToObject("a[fruits][red][]=strawberry&a[1]=five&a[fruits][red][]=cherry&a[fruits][yellow][]=lemon&a[fruits][yellow][688]=banana"));
这将输出如下内容:
{ a: '1' }
{ a: '3' }
{ a: { '0': '1' } }
{ a: { '0': '1', '1': 'pomme' } }
{ a: { '0': 'one', '1': 'five' } }
{ foo: 'bar', number: '1234' }
{
a: { fruits: { red: { '0': 'strawberry' } } }
}
{
a: {
'1': 'five',
fruits: {
red: { '0': 'strawberry', '1': 'cherry' },
yellow: { '0': 'lemon', '688': 'banana' }
}
}
}
注意:它基于loctus parse_str函数(https://locutus.io/php/strings/parse_str/)。
最简单的方法之一是使用URLSearchParam接口。
下面是工作代码片段:
let paramObj={},
querystring=window.location.search,
searchParams = new URLSearchParams(querystring);
//*** :loop to add key and values to the param object.
searchParams.forEach(function(value, key) {
paramObj[key] = value;
});
一个简单的答案是内置本地Node模块。(没有第三方npm模块)
querystring模块提供了解析和格式化URL查询字符串的实用程序。可以使用以下方法访问:
const querystring = require('querystring');
const body = "abc=foo&def=%5Basf%5D&xyz=5"
const parseJSON = querystring.parse(body);
console.log(parseJSON);
/** * 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
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; }
ES6有一个非常简单而不正确的答案:
console.log(
Object.fromEntries(new URLSearchParams(`abc=foo&def=%5Basf%5D&xyz=5`))
);
但是这一行代码不包括多个相同的键，你必须使用更复杂的东西:
function parseParams(params) {
const output = [];
const searchParams = new URLSearchParams(params);
// Set will return only unique keys()
new Set([...searchParams.keys()])
.forEach(key => {
output[key] = searchParams.getAll(key).length > 1 ?
searchParams.getAll(key) : // get multiple values
searchParams.get(key); // get single value
});
return output;
}
console.log(
parseParams('abc=foo&cars=Ford&cars=BMW&cars=Skoda&cars=Mercedes')
)
代码将生成如下结构:
[
abc: "foo"
cars: ["Ford", "BMW", "Skoda", "Mercedes"]
]
许多其他的解决方案没有考虑到边界情况。
这个可以处理
空键a=1&b=2& 空值a=1&b 空值a=1&b= 未编码的等号a=1&b=2=3=4
decodeQueryString: qs => {
// expects qs to not have a ?
// return if empty qs
if (qs === '') return {};
return qs.split('&').reduce((acc, pair) => {
// skip no param at all a=1&b=2&
if (pair.length === 0) return acc;
const parts = pair.split('=');
// fix params without value
if (parts.length === 1) parts[1] = '';
// for value handle multiple unencoded = signs
const key = decodeURIComponent(parts[0]);
const value = decodeURIComponent(parts.slice(1).join('='));
acc[key] = value;
return acc;
}, {});
},
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