我在我的PhoneGap應用這個問題。好像如果其他方法無法正常工作在這裏。驗證登錄
這是我的代碼:
function onSuccess(position){
$("#Login").click(function(evt){
var username = $("#username").val();
var password = $("#password").val();
var d = new Date();
var date = d.getUTCDate();
var hour = d.getUTCHours()+8;
var minutes = d.getUTCMinutes();
var secs = d.getUTCSeconds();
var year = d.getUTCFullYear();
var mon = d.getUTCMonth() + 1;
var day = d.getUTCDay();
var time = year + "-" + mon + "-" + date + " " + hour + ":" + minutes + ":" + secs;
var sendData = {"username": username, "password" : password, "time" : time};
$.ajax({
type: "POST",
url: "http://192.168.254.107/webs/main/ajax/validateLogin.php",
data: sendData,
success: function(data) {
$("#info").html(data);
var ReturnMessage = data; // I also tried data.message here but it gives me "undefined"
if (ReturnMessage != "" && ReturnMessage != "Invalid Login!") {
alert(ReturnMessage);
localStorage.setItem("message", ReturnMessage);
document.location.href = "trackme.html";
} else {
alert(ReturnMessage);
}
}
});
});
}
什麼都ReturnMessage的價值是,它總是這樣的,如果條件的方法。如果我得到了「在成功的記錄!」從我的PHP文件重定向到trackme.html消息,如果我得到「無效登錄!」,這也重定向到trackme.html這意味着如果其他條件不正常。感謝您的幫助
嗨,我在哪裏去的地方JSON.stringify(ReturnMessage)?我需要替換那裏的代碼嗎?謝謝。 – miles