我有一個AJAX功能,更新我的數據庫。該功能很完善和更新數據庫後,我打電話我已經創建了successAlert()函數。但是現在我想在錯誤的情況下調用錯誤函數,但是在故意破壞代碼的情況下,我仍然得到successAlert()。阿賈克斯返回錯誤調用錯誤AJAX調用錯誤
的Ajax/JavaScript的:
var share = "test"
var custid = "test"
$.ajax({
url: "assets/ajax/customer-rec.php",
type: "POST",
data: {UpdateAccount: "yes",custid: custid,share: share},
success: function(result){
successAlert()
},
error: function(result){
errorAlert()
}
});
PHP更新數據庫
if (isset($_POST['UpdateAccount'])){
$custid = $_POST['custid'];
$share = $_POST['share'];
$query="UPDATE `users` SET `share_ord`='$share' WHERE id= $custid";
$stmt = mysql_query($query);
if($stmt === false){
return false
}
}
你是如何創建一個失敗?你有沒有重新命名「assets/ajax/customer-rec.php」? –