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
}
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
}
当前回答
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;
}
}
}
其他回答
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: ""}
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。
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;
}, {});
},
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);
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;
}