Javascript 如何将x-www-form-urlencoded字符串转换为JSON?
应用程序示例/x-www-form-urlencoded stringJavascript 如何将x-www-form-urlencoded字符串转换为JSON?,javascript,json,Javascript,Json,应用程序示例/x-www-form-urlencoded string CorrelationId=1&PickedNumbers%5B%5D=1&PickedNumbers%5B%5D=2&PickedNumbers%5B%5D=3&PickedNumbers%5B%5D=4 转换为JSON var gamePlayData = { CorrelationId: gameId, PickedNumbers: ["
CorrelationId=1&PickedNumbers%5B%5D=1&PickedNumbers%5B%5D=2&PickedNumbers%5B%5D=3&PickedNumbers%5B%5D=4
转换为JSON
var gamePlayData = {
CorrelationId: gameId,
PickedNumbers: ["1","2","3","4"]
};
试试这个->
// convert string to object
str = 'a=6&id=99';
var arr = str.split('&');
var obj = {};
for(var i = 0; i < arr.length; i++) {
var bits = arr[i].split('=');
obj[bits[0]] = bits[1];
}
//alert(obj.a);
//alert(obj.id);
// convert object back to string
str = '';
for(key in obj) {
str += key + '=' + obj[key] + '&';
}
str = str.slice(0, str.length - 1);
alert(str);
//将字符串转换为对象
str='a=6&id=99';
var arr=str.split('&');
var obj={};
对于(变量i=0;i
或者使用这个(JQuery)以下代码应该可以实现这一点:
var str = 'CorrelationId=1&PickedNumbers%5B%5D=1&PickedNumbers%5B%5D=2&PickedNumbers%5B%5D=3&PickedNumbers%5B%5D=4';
var keyValuePairs = str.split('&');
var json = {};
for(var i=0,len = keyValuePairs.length,tmp,key,value;i <len;i++) {
tmp = keyValuePairs[i].split('=');
key = decodeURIComponent(tmp[0]);
value = decodeURIComponent(tmp[1]);
if(key.search(/\[\]$/) != -1) {
tmp = key.replace(/\[\]$/,'');
json[tmp] = json[tmp] || [];
json[tmp].push(value);
}
else {
json[key] = value;
}
}
var str='CorrelationId=1&PickedNumbers%5B%5D=1&PickedNumbers%5B%5D=2&PickedNumbers%5B%5D=3&PickedNumbers%5B%5D=4';
var keyValuePairs=str.split('&');
var json={};
对于(var i=0,len=keyValuePairs.length,tmp,key,value;i,这里有一个纯JavaScript的方法。JavaScript框架也可以帮助您解决这个问题。EDIT:为了方便起见,我也加入了字典解析。参见第二个示例
function decodeFormParams(params) {
var pairs = params.split('&'),
result = {};
for (var i = 0; i < pairs.length; i++) {
var pair = pairs[i].split('='),
key = decodeURIComponent(pair[0]),
value = decodeURIComponent(pair[1]),
isArray = /\[\]$/.test(key),
dictMatch = key.match(/^(.+)\[([^\]]+)\]$/);
if (dictMatch) {
key = dictMatch[1];
var subkey = dictMatch[2];
result[key] = result[key] || {};
result[key][subkey] = value;
} else if (isArray) {
key = key.substring(0, key.length-2);
result[key] = result[key] || [];
result[key].push(value);
} else {
result[key] = value;
}
}
return result;
}
decodeFormParams("CorrelationId=1&PickedNumbers%5B%5D=1&PickedNumbers%5B%5D=2&PickedNumbers%5B%5D=3&PickedNumbers%5B%5D=4");
// => {"CorrelationId":"1","PickedNumbers":["1","2","3","4"]}
decodeFormParams("a%5Bb%5D=c&a%5Bd%5D=e");
// => {"a":{"b":"c","d":"e"}}
函数decodeFormParams(params){
变量对=参数拆分('&'),
结果={};
对于(变量i=0;i{“CorrelationId”:“1”,“PickedNumber”:[“1”,“2”,“3”,“4”]}
decodeFormParams(“a%5Bb%5D=c&a%5Bd%5D=e”);
//=>{“a”:{“b”:“c”,“d”:“e”}
我最近一直在处理这个问题:我必须解析可能包含嵌套多达5层的对象的数据。我需要代码来处理这两个相当复杂的数据,但不能不解码像id=213这样简单的URI
我花了相当长的时间在谷歌上,试图找到一个(半)优雅的解决方案来解决这个问题,这个问题不断出现。因为它每天有1个浏览量(给予或接受),我决定将我的解决方案发布在这里,希望它能帮助别人:
function form2Json(str)
{
"use strict";
var obj,i,pt,keys,j,ev;
if (typeof form2Json.br !== 'function')
{
form2Json.br = function(repl)
{
if (repl.indexOf(']') !== -1)
{
return repl.replace(/\](.+?)(,|$)/g,function($1,$2,$3)
{
return form2Json.br($2+'}'+$3);
});
}
return repl;
};
}
str = '{"'+(str.indexOf('%') !== -1 ? decodeURI(str) : str)+'"}';
obj = str.replace(/\=/g,'":"').replace(/&/g,'","').replace(/\[/g,'":{"');
obj = JSON.parse(obj.replace(/\](.+?)(,|$)/g,function($1,$2,$3){ return form2Json.br($2+'}'+$3);}));
pt = ('&'+str).replace(/(\[|\]|\=)/g,'"$1"').replace(/\]"+/g,']').replace(/&([^\[\=]+?)(\[|\=)/g,'"&["$1]$2');
pt = (pt + '"').replace(/^"&/,'').split('&');
for (i=0;i<pt.length;i++)
{
ev = obj;
keys = pt[i].match(/(?!:(\["))([^"]+?)(?=("\]))/g);
for (j=0;j<keys.length;j++)
{
if (!ev.hasOwnProperty(keys[j]))
{
if (keys.length > (j + 1))
{
ev[keys[j]] = {};
}
else
{
ev[keys[j]] = pt[i].split('=')[1].replace(/"/g,'');
break;
}
}
ev = ev[keys[j]];
}
}
return obj;
}
它整洁地返回一个对象,当通过JSON.stringify
时,该对象如下所示:
{"id":"007","name":{"title":"agent","first":"james","last":"bond"},"personalia":{"weakness":"women","occupation":"spy","strength":"women"},"tools":{"movement":{"far":"DBS","slow":"foot"},"weapons":{"close":{"silent":"garrot"},"medium":{"silent":"pistol_supressed","loud":"smg"},"far":{"silent":"sniper"}}}}
它通过JSLICT检查,忽略空白,<代码> <代码> >代码> [^…] /代码>,接受<代码> ++>代码>。总之,我认为这是可以接受的。 您需要jQueRe.PARAM。选项之一是如果您使用节点或BurSeriType,可以使用。
var qs = require('qs')
var encodedString = "CorrelationId=1&PickedNumbers%5B%5D=1&PickedNumbers%5B%5D=2&PickedNumbers%5B%5D=3&PickedNumbers%5B%5D=4"
console.log(qs.parse(encodedString))
// { CorrelationId: '1', PickedNumbers: [ '1', '2', '3', '4' ] }
这是Node.js现在的核心模块:
也适用于编码字符:
var json2 = qs.parse('http%3A%2F%2Fexample.com&sad=salad')
// { url: 'http://example.com', sad: 'salad' }
string str=“RESULT=0&PNREF=A10AABBF8DF2&RESPMSG=Approved&AUTHCODE=668PNI&PREFPSMSG=No Rules Triggered&postfspsmsg=No Rules Triggered”
var sr=str.Replace(&),“=”;
字符串[]sp=sr.Split('=');
var spl=sp.长度;
int n=1;
var ss=“{”;
对于(var k=0;k
一行:
s='a=1&b=2&c=3';
Object.fromEntries(
s、 拆分(“&”)
.map(s=>s.split('='))
.map(pair=>pair.map(decodeURIComponent)))
//->{a:“1”,b:“2”,c:“3”}
如果要将重复参数表示为数组:
s='a=1&b=2&c[]=3&c[]=4&c[]=5&c[]=6';
s
.split(“&”)
.map(s=>s.split('='))
.map(pair=>pair.map(decodeURIComponent))
.reduce((备注,[键,值])=>{
如果(!(输入备忘录)){memo[key]=value;}
否则{
if(!(memo[key]数组实例))
备注[键]=[备注[键],值];
其他的
备注[键]。推送(值);
}
返回备忘录;
}, {})
//->{“a”:“1”、“b”:“2”、“c[]”:[“3”、“4”、“5”、“6”]}
jQuery.param进行序列化-问题是关于反向操作。在节点中,您可以使用内置的querystring
如果您添加一些解释会更好。我正在获取引用错误:如果(typeof form2Json.br!=“function”)您是最好的,则在第行中未定义form2Json。:)如果数据包含方括号,甚至引号,则代码无法正常工作代码>将导致错误请使用真实变量名作为示例。这允许其他人阅读您答案中的代码。它是否考虑嵌套数组?
var qs = require('querystring')
var json = qs.parse('why=not&sad=salad')
// { why: 'not', sad: 'salad' }
var json2 = qs.parse('http%3A%2F%2Fexample.com&sad=salad')
// { url: 'http://example.com', sad: 'salad' }
public static void Main()
{
var sr = str.Replace("&", "=");
string[] sp = sr.Split('=');
var spl = sp.Length;
int n = 1;
var ss = "{";
for (var k = 0; k < spl; k++)
{
if (n % 2 == 0)
{
if (n == spl)
{
ss += '"' + sp[k] + '"';
}
else
{
ss += '"' + sp[k] + '"' + ",";
}
}
else
{
ss += '"' + sp[k] + '"' + ":";
}
n++;
}
ss += "}";
Console.WriteLine(ss);
}