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編輯
感謝，韋斯CI現在有這樣的AJAX代碼：允許登錄表單處理同一個頁面上的PHP腳本而無需刷新頁面
<script type="text/javascript">
$(document).ready(function(){
$("form[name^='login']").submit(function(event) {
event.preventDefault();
var dataToSend = 'username='+$("input[name^='username']").val()+'&password='+$("input[name^='password']").val();
$.ajax({
type: "POST",
url: "index.php",
data: dataToSend,
success: function(response){
if(response == "REDIRECT")
{
window.location = "business_profiles/myReviews.php";
else
{
$("#my_error_div").html(response);
}
}
});
});
});
</script>
但現在，而不是顯示錯誤消息時例如鍵入你的密碼框。無論表單輸入是什麼，我收到的所有信息都是一個提示框，顯示「成功」。
好吧，我試過並試圖找出我自己的。以及聯繫到其他編碼人員，當然搜索stackoverflow的答案。沒有一個看起來與我的情況相符。
我有一個簡單的登錄表單，像這樣：
<form name="login" action="index.php" method="post">
<ul>
<li>
<input type="text" name="username" autocomplete="off" placeholder="Username" autofocus="true"></li>
<li>
<input type="password" name="password" autocomplete="off" placeholder="Password"></li>
<li>
<input type="submit" name="submit" value="Login"> <a href="register.php" id="button">Register</a> <a href="forgot.php" id="button">Forgot Password</a> <a href = "javascript:void(0)" onclick = "document.getElementById('light').style.display='none';document.getElementById('fade').style.display='none'" id="button">Close</a>
</li>
</ul>
</form>
這種形式提交到它是在同一頁上。 （PHP腳本是正確的登錄表單之上。），這是如下：
<?php
//If the user has submitted the form
if($_POST['username']){
//protect the posted value then store them to variables
$username = protect($_POST['username']);
$password = protect($_POST['password']);
//Check if the username or password boxes were not filled in
if(!$username || !$password){
//if not display an error message
echo "<center>You need to fill in a <b>Username</b> and a <b>Password</b>!</center>";
}else{
//if the were continue checking
//select all rows from the table where the username matches the one entered by the user
$res = mysql_query("SELECT * FROM `users` WHERE `username` = '".$username."'");
$num = mysql_num_rows($res);
//check if there was not a match
if($num == 0){
//if not display an error message
echo "<center>The <b>Username</b> you supplied does not exist!</center>";
}else{
//if there was a match continue checking
//select all rows where the username and password match the ones submitted by the user
$res = mysql_query("SELECT * FROM `users` WHERE `username` = '".$username."' AND `password` = '".$password."'");
$num = mysql_num_rows($res);
//check if there was not a match
if($num == 0){
//if not display error message
echo "<center>The <b>Password</b> you supplied does not match the one for that username!</center>";
}else{
//if there was continue checking
//split all fields fom the correct row into an associative array
$row = mysql_fetch_assoc($res);
//check to see if the user has not activated their account yet
if($row['active'] != 1){
//if not display error message
echo "<center>You have not yet <b>Activated</b> your account!</center>";
}else{
//if they have log them in
//set the login session storing there id - we use this to see if they are logged in or not
$_SESSION['uid'] = $row['id'];
//show message
echo "<center>You have successfully logged in!</center>";
//update the online field to 50 seconds into the future
$time = date('U')+50;
mysql_query("UPDATE `users` SET `online` = '".$time."' WHERE `id` = '".$_SESSION['uid']."'");
//redirect them to the usersonline page
echo 'REDIRECT';
exit;
}
}
}
}
}
?>
我需要的形式處理腳本而無需刷新頁面。我的登錄表單位於燈箱內，因此如果在頁面刷新燈箱時隱藏了無效密碼等錯誤，則必須再次單擊登錄以查明您做錯了什麼。我只是希望表單在不刷新頁面的情況下處理php腳本，因此燈箱永遠不會隱藏，直到用戶成功登錄。然後用戶被重定向到他們的配置文件。
你的意思是像AJAX顯示頁面？ – intelis 2013-03-28 04:00:47
對於這個場景使用'AJAX'。 – 2013-03-28 04:01:55
因此，我將不得不將腳本的所有編碼更改爲AJAX？ – user2109152 2013-03-28 04:03:26