我想顯示錯誤,如果用戶名和密碼錯了,我想從昨天開始。不知道什麼wrong.every時控去錯誤的,並顯示error.tried json.but jQuery的AJAX不是working.i一個新的,返回json_encode錯誤從PHP到jquery
<script>
$(document).ready(function() {
$("#LoginForm").submit(function(e) {
$("#simple-msg1").html("<img src='img/loading.gif'/>");
var postData ="";
postData = $('#LoginForm').serializeArray();
var formURL = $('#LoginForm').attr("action");
$.ajax({
url : formURL,
type: "POST",
data : postData,
dataType:'json',
success:function(data, textStatus, jqXHR) {
alert(data.error);
if(data.error == 1) {
$("#simple-msg1").html('<pre><code class="prettyprint">'+data.message+'</code>< /pre>');
} else {
$("#simple-msg1").html('<pre><code class="prettyprint"> Login Successfull </code></pre>');
window.location = "/property/Dealer/ManageProfile.php?Login=successfull";
}
},
error:function(data,textStatus) {
$("#simple-msg1").html('<pre><code class="prettyprint"> wrong username or password 11</code></pre>');
}
});
e.preventDefault(); //STOP default action
});
$("#Button1").click(function() {
$("#LoginForm").submit(); //SUBMIT FORM
});
});
</script>
這是PHP文件
<?php
ob_start();
if(session_id() == '')
{
session_start();
}
include 'config.php';
$error = '0';
$message = 'Valid';
$redirect = 'Dealer/EditLoginDetails.php';
$myusername=$_POST['txtusername'];
$mypassword=$_POST['txtpassword'];
$myusername = stripslashes($myusername);
$mypassword = stripslashes($mypassword);
$myusername = mysql_real_escape_string($myusername);
$mypassword=mysql_real_escape_string($mypassword);
$qry = "SELECT UserName,Type_user FROM login WHERE UserName = '".$myusername."' AND password = '".$mypassword."' ";
$result = mysql_query($qry) or die ("Query failed");
$UserData = mysql_fetch_array($result);
if($UserData['UserName'] != "") {
//echo $UserData['UserName'];
$_SESSION['UserId'] = $myusername;
$typ = $UserData['Type_user'];
if ($typ == "Dealer") {
header('location:/Dealer/EditLoginDetails.php');
//echo "dealer";
//echo json_encode(array('success'=>'true'));
//header('location:/Dealer/EditLoginDetails.php');
} else if ($typ == "Individual") {
header('location:/Dealer/EditLoginDetails.php');
} else {
header('location:/Builder/managep.php');
}
} else {
$error = '1';
$message = 'Invalid username or password';
// echo "forbiddnt";
//header('HTTP/1.0 403 Forbidden');
//echo " wrong username or password";
}
echo json_encode(array('error' => $error, 'message' => $message, 'redirect' => $redirect));
?>
如果使用'.serialize()'而不是'.serializeArray()',它會起作用嗎? – Barmar
我還沒有試過這個 – dvirus
試過了不行:) :( – dvirus