[英]Return value from php to ajax success and error
我正在通過參考教程來使用Jquery mobile和php進行用戶身份驗證。 但是本教程的后端是mongoDB。 我希望我的應用程序連接到PHP和mysql。
我對教程代碼進行了一些更改,我對php返回錯誤或成功感到困惑。
AJAX代碼
$.ajax({
type: 'POST',
url: BookIt.Settings.signUpUrl,
data:"submitted=" + "1" + "&mobile=" + mobileNumber + "&firstName=" + firstName + "&lastName=" + lastName + "&password=" + password,
dataType: "json",
success: function (resp) {
alert(resp);
console.log("success");
if (resp.success === true) {
$.mobile.navigate("signup-succeeded.html");
return;
}
if (resp.extras.msg) {
switch (resp.extras.msg) {
case BookIt.ApiMessages.DB_ERROR:
case BookIt.ApiMessages.COULD_NOT_CREATE_USER:
// TODO: Use a friendlier error message below.
$ctnErr.html("<p>Oops! BookIt had a problem and could not 2 register you. Please try again in a few minutes.</p>");
$ctnErr.addClass("bi-ctn-err").slideDown();
break;
case BookIt.ApiMessages.MOBILENUMBER_ALREADY_EXISTS:
$ctnErr.html("<p>The mobile number that you provided is already registered.</p>");
$ctnErr.addClass("bi-ctn-err").slideDown();
$txtMobileNumber.addClass(invalidInputStyle);
break;
}
}
},
error: function (e) {
console.log(e.message);
// TODO: Use a friendlier error message below.
$ctnErr.html("<p>Oops! BookIt had a problem and could not register you. Please try again in a few minutes.</p>");
$ctnErr.addClass("bi-ctn-err").slideDown();
}
});
MY PHP返回JSON編碼輸出
echo json_encode("success");
API消息
var BookIt = BookIt || {};
BookIt.ApiMessages = BookIt.ApiMessages || {};
BookIt.ApiMessages.EMAIL_NOT_FOUND = 0;
BookIt.ApiMessages.INVALID_PWD = 1;
BookIt.ApiMessages.DB_ERROR = 2;
BookIt.ApiMessages.NOT_FOUND = 3;
BookIt.ApiMessages.EMAIL_ALREADY_EXISTS = 4;
BookIt.ApiMessages.COULD_NOT_CREATE_USER = 5;
BookIt.ApiMessages.PASSWORD_RESET_EXPIRED = 6;
BookIt.ApiMessages.PASSWORD_RESET_HASH_MISMATCH = 7;
BookIt.ApiMessages.PASSWORD_RESET_EMAIL_MISMATCH = 8;
BookIt.ApiMessages.COULD_NOT_RESET_PASSWORD = 9;
BookIt.ApiMessages.PASSWORD_CONFIRM_MISMATCH = 10;
BookIt.ApiMessages.MOBILENUMBER_ALREADY_EXISTS=11;
BookIt.ApiMessages.USERNAME_ALREADY_EXISTS=12;
我如何將錯誤/成功返回到此ajax,所以它工作正常
您正在檢查
if (resp.success === true)
這意味着json數據必須采用以下格式:
{"success":true}
由以下人員產生:
echo json_encode( ['success'=>true] ); // or array() instead of [] for PHP < 5.4
同樣,對於錯誤( if (resp.extras.msg)
),
echo json_encode( [ 'extras' => ['msg' => $errorcode] ] );
產生此JSON($ errorcode = 2(BookIt.ApiMessages.DB_ERROR))
{"extras":{"msg":2}}
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