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+"&open>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');