我有一個使用JQuery和PHP驗證用戶名和密碼的登錄表單。我正嘗試從擴展mysql切換到mysqli以獲得更好的練習,並且遇到了很多麻煩。我認爲這個錯誤存在於JQuery中,因爲我已經測試了login.php驗證,並且它工作正常。使用PHP和JQuery驗證來自MySQL數據庫的登錄信息登錄表格
因此,這裏是我的代碼:
的index.php:
<!doctype html>
<html>
<head>
<script src="//ajax.googleapis.com/ajax/libs/jquery/2.0.0/jquery.min.js"></script>
<script type="text/javascript">
$(document).ready(function(){
$("#login_a").click(function(){
$("#shadow").fadeIn("normal");
$("#login_form").fadeIn("normal");
$("#user_name").focus();
});
$("#cancel_hide").click(function(){
$("#login_form").fadeOut("normal");
$("#shadow").fadeOut();
});
$("#login").click(function(){
var username=$('#user_name').val();
var password=$('#password').val();
$.ajax({
type: "POST",
url: "login.php",
data: "name="+username+"&pwd="+password,
success: function(html){
if(html=='true'){
$("#login_form").fadeOut("normal");
$("#shadow").fadeOut();
$("#profile").html("<a href='logout.php' id='logout'>Logout</a>");
}
else{
$("#add_err").html("*Wrong username or password");
}
},
beforeSend:function(){
$("#add_err").html("Loading...")
}
});
return false;
});
});
</script>
</head>
<body>
<?php session_start(); ?>
<div id="profile">
<?php if(isset($_SESSION['user_name'])){ ?>
<a href='logout_script_2.php' id='logout'>Logout</a>
<?php }else {?>
<a id="login_a" href="#">login</a>
<?php } ?>
</div>
<div id="login_form">
<form action="login_script_2.php" method="POST">
<label>User Name:</label>
<input type="text" id="user_name" name="user_name" />
<label>Password:</label>
<input type="password" id="password" name="password" />
<label></label><br/>
<input type="submit" id="login" value="Login" />
<input type="button" id="cancel_hide" value="Cancel" />
</form>
<div class="err" id="add_err"><br></div>
</div>
<div id="shadow" class="popup"></div>
</body>
</html>
的login.php
<?php
session_start();
$con = mysqli_connect("localhost","root","PW","db") or die("Connection error: " . mysqli_error($con));
if(isset($_POST['submit'])){
$username = $_POST['name'];
$password = $_POST['pwd'];
$stmt = $con->prepare("SELECT * FROM tapp_login WHERE username = ? AND password = ? LIMIT 1");
$stmt->bind_param('ss', $username, $password);
$stmt->execute();
$stmt->bind_result($username, $password);
$stmt->store_result();
if($stmt->num_rows == 1)
{
if($stmt->fetch())
{
$_SESSION['Logged'] = 1;
$_SESSION['user_name'] = $username;
echo 'Access granted';
exit();
}
}
else {
echo "*Wrong username or password";
}
$stmt->close();
}
else {
}
$con->close();
?>
logout.php
<?php
session_start();
unset($_SESSION['user_name']);
header('Location: index.php');
?>
所有運行的嘗試代碼給了我JQuer中的錯誤y驗證「*錯誤的用戶名或密碼」。先謝謝了!
感謝您的提示。我嘗試了兩種方法,仍然得到相同的錯誤。任何其他想法? – Rob
你檢查了你的控制檯日誌,看看你的成功返回什麼 - 'html'? – Sean
我不知道該怎麼做,請您詳細說明一下嗎? – Rob