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
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,"\":\"")) + '"}')
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》。
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}， {})
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;
}
}
}
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