const _ = require('lodash');
const url = `https://myWeb.com/abc/def/?xyz=sdfgkjhg&myName=javascript&status=married&sex=male&age=25`;
const queryParam = url.split('&');
let start = _.filter(queryParam,(param) =>{
return _.includes(param,'start');
});
startIndex = _.split(startIndex, '=')[1];
console.log("=========="+start);
发布于 2017-08-07 14:36:06
使用url库
var urlPackage = require('url');
var url_parts = urlPackage.parse(url, true);
var query = url_parts.query;现在您可以访问任何varaible查询。例如
console.log(query.myName); // it will print javascript发布于 2017-08-07 15:01:29
var query = 'https://myWeb.com/abc/def/?xyz=sdfgkjhg&myName=javascript&status=married&sex=male&age=25'.split('?')[1].split('&')
var final = query.reduce((m,p)=>{
p = p.split('=');
m[p[0]] = p[1]
return m
},{})
console.log('***',final); 发布于 2017-08-08 06:55:52
您可以使用RegExp解析url,将结果Array#map到一个由参数名称和参数值对组成的数组,并从该数组创建一个Map。现在,您可以使用paramsMap.get('paramName')从映射中获取任何参数值。
const url = `https://myWeb.com/abc/def/?xyz=sdfgkjhg&myName=javascript&status=married&sex=male&age=25`;
const paramsMap = new Map(url.match(/[^?&=]+=[^&=]+/g).map(s => s.split('=')));
console.log(paramsMap.get('status'));
https://stackoverflow.com/questions/45540347
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