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
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 }, {}))
其他回答
//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
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}， {})
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' }
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);
