js頁面跳轉傳參

在需要跳轉得頁面通過url拼接需要傳遞得參數

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

跳轉后得頁面用getQueryString方法獲取參數

 

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');