js页面跳转传参

在需要跳转得页面通过url拼接需要传递得参数

window.location.href = "../html/trainingOrghome.html?" + "accountTypeId="+2+"&openid="+openid+"&nickname="+nickname+"&headimgurl="+headimgurl 

跳转后得页面用getQueryString方法获取参数

 

function getQueryString(name) { console.log('.......') var reg = new RegExp("(^|&)" + name + "=([^&]*)(&|$)", "i"); var r = window.location.search.substr(1).match(reg); if (r != null && r[2] != "false") return unescape(r[2]); return false; } let openID = getQueryString('openid'); let accountTypeId = getQueryString('accountTypeId'); let nickname = getQueryString('nickname'); let headimgurl = getQueryString('headimgurl');