我有一个这样的字符串:
abc=foo&def=%5Basf%5D&xyz=5
如何将其转换为像这样的JavaScript对象?
{
abc: 'foo',
def: '[asf]',
xyz: 5
}
答案 0 :(得分:278)
此编辑会根据评论改进并解释答案。
var search = location.search.substring(1);
JSON.parse('{"' + decodeURI(search).replace(/"/g, '\\"').replace(/&/g, '","').replace(/=/g,'":"') + '"}')
示例强>
分五个步骤解析abc=foo&def=%5Basf%5D&xyz=5
:
abc=foo","def=[asf]","xyz=5
abc":"foo","def":"[asf]","xyz":"5
{"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包含更多字符。它使用reviver函数进行URI解码:
var search = location.search.substring(1);
JSON.parse('{"' + search.replace(/&/g, '","').replace(/=/g,'":"') + '"}', function(key, value) { return key===""?value:decodeURIComponent(value) })
示例
search = "abc=foo&def=%5Basf%5D&xyz=5&foo=b%3Dar";
给出
Object {abc: "foo", def: "[asf]", xyz: "5", foo: "b=ar"}
单行:
JSON.parse('{"' + decodeURI("abc=foo&def=%5Basf%5D&xyz=5".replace(/&/g, "\",\"").replace(/=/g,"\":\"")) + '"}')
答案 1 :(得分:32)
从ES6开始,Javascript提供了几种结构来创建针对此问题的高性能解决方案。
这包括使用URLSearchParams和iterators
let params = new URLSearchParams('abc=foo&def=%5Basf%5D&xyz=5');
params.get("abc"); // "foo"
如果您的用例需要您将其实际转换为对象,则可以实现以下功能:
function paramsToObject(entries) {
let result = {}
for(let entry of entries) { // each 'entry' is a [key, value] tupple
const [key, value] = entry;
result[key] = value;
}
return result;
}
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"}
我们可以使用Object.fromEntries(目前处于第4阶段),将paramsToObject
替换为Object.fromEntries(entries)
。
由于URLParams返回iterable对象,所以使用spread operator而不是调用.entries
也会按照其规范生成条目:
要迭代的值对是列表名称-值对,其中 键是名称,值是值。
const urlParams = new URLSearchParams('abc=foo&def=%5Basf%5D&xyz=5');
const params = Object.fromEntries(...params); // {abc: "foo", def: "[asf]", xyz: "5"}
注意:根据URLSearchParams spec
,所有值都是自动字符串答案 2 :(得分:24)
在&
上拆分以获取名称/值对,然后在=
上拆分每对。这是一个例子:
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;
}, {});
使用正则表达式的另一种方法:
var obj = {};
str.replace(/([^=&]+)=([^&]*)/g, function(m, key, value) {
obj[decodeURIComponent(key)] = decodeURIComponent(value);
});
这是改编自John Resig的"Search and Don’t Replace"。
答案 3 :(得分:13)
这是简单版本,显然你想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]);
}
答案 4 :(得分:10)
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documentation。
答案 5 :(得分:9)
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);
return ({ ...obj, [key]: value })
}, {});
答案 6 :(得分:8)
另一种基于最新标准的URLSearchParams(https://developer.mozilla.org/en-US/docs/Web/API/URLSearchParams)
的解决方案function getQueryParamsObject() {
const searchParams = new URLSearchParams(location.search.slice(1));
return searchParams
? _.fromPairs(Array.from(searchParams.entries()))
: {};
}
请注意,此解决方案正在使用
Array.from (https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/from)
为了简单起见,lodash的和 _。fromPairs (https://lodash.com/docs#fromPairs)。
创建更兼容的解决方案应该很容易,因为您可以访问 searchParams.entries()迭代器。
答案 7 :(得分:7)
到目前为止我找到的建议解决方案并未涉及更复杂的情况。
我需要转换像
这样的查询字符串 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
进入像:
这样的对象{
"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
INTO:
{
"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"
}
}
}
CODE:
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;
};
答案 8 :(得分:6)
我遇到了同样的问题,在这里尝试了解决方案,但是没有一个真正有用,因为我在URL参数中有数组,如下所示:
?param[]=5¶m[]=8&othr_param=abc¶m[]=string
所以我最终编写了自己的JS函数,它在URI中创建了一个数组:
/**
* 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;
}
答案 9 :(得分:5)
ES6一个衬板(如果看到长线的话我们可以这样称呼)
[...new URLSearchParams(location.search).entries()].reduce((prev, [key,val]) => {prev[key] = val; return prev}, {})
答案 10 :(得分:5)
ES6单线。干净简单。
var obj = [...new URLSearchParams(location.search).entries()].reduce((sum, [key,val]) => Object.assign({[key]:val}, sum), {});
答案 11 :(得分:5)
使用URLSearchParam接口执行此操作的最简单方法之一。
下面是工作代码段:
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;
});
答案 12 :(得分:5)
使用ES6,URL API和URLSearchParams API。
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;
}
答案 13 :(得分:2)
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;
}
}
}
答案 14 :(得分:2)
我没有意识到的原生解决方案。如果您偶然使用该框架,Dojo有一个内置的反序列化方法。
否则你可以自己实现它:
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:添加了decodeURIComponent()
答案 15 :(得分:2)
使用URLSearchParams
JavaScript Web API
var paramsString = "q=forum&topic=api";
//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 = {q:"forum",topic:"api"}
//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());
注意: IE不支持
答案 16 :(得分:2)
使用本机节点模块构建的一个简单答案。(无第三方npm模块)
querystring模块提供用于解析和格式化URL查询字符串的实用程序。可以使用以下方式访问它:
const querystring = require('querystring');
const body = "abc=foo&def=%5Basf%5D&xyz=5"
const parseJSON = querystring.parse(body);
console.log(parseJSON);
答案 17 :(得分:2)
/**
* 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
答案 18 :(得分:2)
有一个名为YouAreI.js的轻量级库已经过测试,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' }
答案 19 :(得分:1)
这似乎是最好的解决方案,因为它需要考虑多个相同名称的参数。
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
我后来决定将它转换为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,第一个只接受参数,但jQuery插件将获取整个url并返回序列化参数。
答案 20 :(得分:1)
如果需要递归,则可以使用小型的 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locutus parse_str函数(https://locutus.io/php/strings/parse_str/)。
答案 21 :(得分:1)
console.log(decodeURI('abc=foo&def=%5Basf%5D&xyz=5')
.split('&')
.reduce((result, current) => {
const [key, value] = current.split('=');
result[key] = value;
return result
}, {}))
答案 22 :(得分:1)
ES6 有一个非常简单且不正确的答案:
console.log(
Object.fromEntries(new URLSearchParams(`abc=foo&def=%5Basf%5D&xyz=5`))
);
但是这一行代码没有覆盖多个相同的键,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"]
]
答案 23 :(得分:1)
针对您的具体情况:
Object.fromEntries(new URLSearchParams('abc=foo&def=%5Basf%5D&xyz=5'));
对于更常见的情况,有人想要将查询参数解析为对象:
Object.fromEntries(new URLSearchParams(location.search));
如果您无法使用Object.fromEntries,这也将起作用:
Array.from(new URLSearchParams(window.location.search)).reduce((o, i) => ({ ...o, [i[0]]: i[1] }), {});
答案 24 :(得分:1)
对于Node,您可以使用以下内容:
const querystring = require('querystring');
querystring.parse('abc=foo&def=%5Basf%5D&xyz=5&foo=b%3Dar');
// returns the object
答案 25 :(得分:1)
如果您使用的是URI.js,则可以使用:
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
};
答案 26 :(得分:1)
{{1}}
答案 27 :(得分:1)
这是我的快速和肮脏的版本,基本上它拆分了由&#39;&amp;&#39;分隔的网址参数。到数组元素,然后迭代该数组添加由&#39; =&#39;分隔的键/值对成对象。我使用decodeURIComponent()将编码字符转换为正常的字符串等价物(因此%20变为空格,%26变为&#39;&amp;&#39;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;
}
示例:
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(由于使用了decodeURIComponent()),但除此之外它不是一个糟糕的起点。
答案 28 :(得分:1)
我在这里使用的是:
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,例如
?a=aa&b=bb
Object {a: "aa", b: "bb"}
重复参数,例如
?a=aa&b=bb&c=cc&c=potato
Object {a: "aa", b: "bb", c: ["cc","potato"]}
缺少钥匙,例如
?a=aa&b=bb&=cc
Object {a: "aa", b: "bb"}
缺少值,例如
?a=aa&b=bb&c
Object {a: "aa", b: "bb"}
上述JSON / regex解决方案在这个古怪的网址上抛出语法错误:
?a=aa&b=bb&c=&=dd&e
Object {a: "aa", b: "bb", c: ""}
答案 29 :(得分:0)
我以这种方式这样做:
const uri = new URL('https://example.org/?myvar1=longValue&myvar2=value')
const result = {}
for (let p of uri.searchParams) {
result[p[0]] = p[1]
}
答案 30 :(得分:0)
在Mike Causer's answer之上构建我已经创建了这个函数,它考虑了具有相同键(foo=bar&foo=baz
)和逗号分隔参数(foo=bar,baz,bin
)的多个参数。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?foo=bar&foo=baz&foo=bez,boz,buz&bar=1,2,3
示例输出
{
foo: ["bar","baz","bez","boz","buz"],
bar: ["1","2","3"]
}
答案 31 :(得分:0)
首先需要定义什么是获取VAR:
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];
}
}
不仅仅是阅读:
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]];
}
}
}
并使用如:
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]];
}
答案 32 :(得分:0)
我还需要在网址(decodeURIComponent doesn't)的查询部分处理+
,因此我改编了沃尔夫冈的代码:
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)}):{};
就我而言,我使用jQuery来获取支持URL的表单参数,然后使用这个技巧来构建一个对象,然后我就可以轻松更新对象上的参数并重建查询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 = $.param(objForm);
答案 33 :(得分:0)
这是一个更精简的 silicakes' approach 版本。
以下函数可以从 USVString
或 Location
解析查询字符串。
/**
* Returns a plain object representation of a URLSearchParams object.
* @param {USVString} search - A URL querystring
* @return {Object} a key-value pair object from a URL querystring
*/
const parseSearch = (search) =>
[...new URLSearchParams(search).entries()]
.reduce((acc, [key, val]) => ({
...acc,
// eslint-disable-next-line no-nested-ternary
[key]: Object.prototype.hasOwnProperty.call(acc, key)
? Array.isArray(acc[key])
? [...acc[key], val]
: [acc[key], val]
: val
}), {});
/**
* Returns a plain object representation of a URLSearchParams object.
* @param {Location} location - Either a document or window location, or React useLocation()
* @return {Object} a key-value pair object from a URL querystring
*/
const parseLocationSearch = (location) => parseSearch(location.search);
console.log(parseSearch('?foo=bar&x=y&ids=%5B1%2C2%2C3%5D&ids=%5B4%2C5%2C6%5D'));
.as-console-wrapper { top: 0; max-height: 100% !important; }
这是上面代码的一行代码(125 字节):
<块引用>其中 f
是 parseSearch
f=s=>[...new URLSearchParams(s).entries()].reduce((a,[k,v])=>({...a,[k]:a[k]?Array.isArray(a[k])?[...a[k],v]:[a[k],v]:v}),{})
这是一种序列化和更新的方法:
const parseSearch = (search) =>
[...new URLSearchParams(search).entries()]
.reduce((acc, [key, val]) => ({
...acc,
// eslint-disable-next-line no-nested-ternary
[key]: Object.prototype.hasOwnProperty.call(acc, key)
? Array.isArray(acc[key])
? [...acc[key], val]
: [acc[key], val]
: val
}), {});
const toQueryString = (params) =>
`?${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; }
答案 34 :(得分:-1)
许多其他解决方案没有考虑边缘情况。
这个手柄
a=1&b=2&
a=1&b
a=1&b=
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;
}, {});
},