我是移動開發新手。我正在嘗試開發一個登錄表單,用於檢查MySQL數據庫的用戶名和密碼。我已經建立了一些代碼。我有兩個問題。第一個是,在提交表單後,在瀏覽器的URL欄中,我可以看到輸入的用戶名和密碼,即使在我的JQuery腳本上使用POST。第二個問題是頁面繼續返回到login.html,而不是main.html。下面是我用PHP,JQuery和MySQL的實際經驗構建的。使用Jquery Mobile,PHP和MySQL的移動應用程序登錄表格
HTML表單
<div id="loginPage" data-role="page">
<div data-role="header">
<h1>App Authentication</h1>
</div>
<div data-role="content">
<div id="landmark-1" data-landmark-id="1">
<form id="loginForm">
<div data-role="fieldcontain" class="ui-hide-label">
<label for="username">Username:</label>
<input type="text" name="username" id="username" value="" placeholder="Username" />
</div>
<div data-role="fieldcontain" class="ui-hide-label">
<label for="password">Password:</label>
<input type="password" name="password" id="password" value="" placeholder="Password" />
</div>
<input type="submit" value="Login" id="submitButton">
</form>
</div>
</div>
<div data-role="footer">
<h4>© Silva's General Services</h4>
</div>
的auth.php
<?php
$mysqli = new mysqli($mysql_hostname,$mysql_user, $mysql_password, $mysql_database);
header("access-control-allow-origin: *");
header("access-control-allow-methods: GET, POST, OPTIONS");
header("access-control-allow-credentials: true");
header("access-control-allow-headers: Content-Type, *");
header("Content-type: application/json");
// Parse the log in form if the user has filled it out and pressed "Log In"
if (isset($_POST["username"]) && isset($_POST["password"])) {
CRYPT_BLOWFISH or die ('No Blowfish found.');
//This string tells crypt to use blowfish for 5 rounds.
$Blowfish_Pre = '$2a$05$';
$Blowfish_End = '$';
$username = mysql_real_escape_string($_POST['username']);
$password = mysql_real_escape_string($_POST['password']);
$sql = "SELECT id, password, salt, username FROM users WHERE username='$username' LIMIT 1";
$result = $mysqli->query($sql) or die($mysqli->error());
$row = mysqli_fetch_assoc($result);
$hashed_pass = crypt($password, $Blowfish_Pre . $row['salt'] . $Blowfish_End);
if ($hashed_pass == $row['password']) {
$response_array['status'] = 'success';
} else {
$response_array['status'] = 'error';
}
echo json_encode($response_array);
}
$mysqli->close();
?>
jQuery腳本
<script>
$('#loginForm').submit(function(){
e.preventDefault();
jQuery.support.cors = true;
$.ajax({
url: 'http://www.test.com/mobile/auth.php',
crossDomain: true,
type: 'post',
data: $("#loginForm").serialize(),
success: function(data){
if(data.status == 'success'){
window.location.href = 'http://www.test.com/mobile/main.html';
}else if(data.status == 'error'){
alert("Authentication Invalid. Please try again!");
return false;
}
}
});
});
</script>
我真的很感激任何幫助。
謝謝。
Hi @BogdanM!這太棒了。這個小小的「e」正在完成這項工作。非常感謝你的回答! –