我在这里找到了一个工作代码，该代码在他登录时返回用户的相应user_id。但是，我需要使用$ .get或$ .. jquery的$ .post不会刷新页面。我阅读了有关使用$ .get和$ .post的帖子，但我似乎不了解它们。
这是我的工作代码。请帮助我申请$。谢谢！
<?php
$login = array
(
0=>array
( 'username'=>'Art',
'password'=>'p@ssw0rd',
'user_id'=>'1'
),
1=>array
( 'username'=>'Berto',
'password'=>'1234',
'user_id'=>'2'
),
2=>array
( 'username'=>'Carrie',
'password'=>'5678',
'user_id'=>'3'
),
3=>array
( 'username'=>'Dino',
'password'=>'qwer',
'user_id'=>'4'
),
4=>array
( 'username'=>'Ely',
'password'=>'asdf',
'user_id'=>'5'
)
);
if (isset($_POST['submit']))
{
$found_user = false;
$uname = $_POST['user'];
$pass = $_POST['password'];
foreach ($login as $sub_array) {
if ($uname == $sub_array['username'] && $pass == $sub_array['password']) {
$found_user = true;
$message = $sub_array['user_id'];
break;
}
}
if ($found_user) {
$message = ( $sub_array['user_id'] );
echo "<script type='text/javascript'>
alert('Your user ID is : $message');
</script>";
} else {
echo "<script type='text/javascript'>
alert('Invalid username/password.');
</script>";
}
}
?>
<body>
<form action="" method="post">
<label for="name"><span class="nameLabel">Username: </span></label>
<span class="nameLabel">
<input name="user" type="text" class="name" id="user" />
</span>
<label for="pass" class="passwordLabel">Password: </label>
<input name="password" type="password" class="password" id="password" />
<input name="submit" type="submit" class="logIn" id="login" value="Log In"/>
</form>
</body>
您需要ajax来满足您的要求：
1）您需要在单独的PHP文件中分开PHP登录功能。
2）单击登录按钮时，您需要进行AJAX调用。您需要为Ajax调用添加一些JavaScript。
3）在您的HTML文件中添加jQuery
您的php文件（我命名为t2.php）
$login = array
(
0=>array
( 'username'=>'Art',
'password'=>'p@ssw0rd',
'user_id'=>'1'
),
1=>array
( 'username'=>'Berto',
'password'=>'1234',
'user_id'=>'2'
),
2=>array
( 'username'=>'Carrie',
'password'=>'5678',
'user_id'=>'3'
),
3=>array
( 'username'=>'Dino',
'password'=>'qwer',
'user_id'=>'4'
),
4=>array
( 'username'=>'Ely',
'password'=>'asdf',
'user_id'=>'5'
)
);
$found_user = false;
$uname = $_POST['user'];
$pass = $_POST['password'];
foreach ($login as $sub_array) {
if ($uname == $sub_array['username'] && $pass == $sub_array['password']) {
$found_user = true;
$message = $sub_array['user_id'];
break;
}
}
if ($found_user) {
$message = ( $sub_array['user_id'] );
echo "Your user ID is : $message";
} else {
echo 'Invalid username/password.';
}
在您的html文件中：
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"></script>
<form action="" method="post" id="login-form">
<label for="name"><span class="nameLabel">Username: </span></label>
<span class="nameLabel">
<input name="user" type="text" class="name" id="user" />
</span>
<label for="pass" class="passwordLabel">Password: </label>
<input name="password" type="password" class="password" id="password" />
<input name="submit" type="button" class="logIn" id="login" value="Log In"/>
</form>
<script type="text/javascript">
$( document ).ready(function() {
$("#login").click(function(e){
e.preventDefault();
$.ajax({
url: "t2.php",
type: "post",
data: $("#login-form").serialize(),
success: function(data){
alert(data);
}
})
})
});
</script>
您需要学习ajax-php-jquery的基础：http：//blog.teamtreehouse.com/beginners-guide-to-to-ajax-development-with-php
with index.php是您的php脚本名称。
$("#login").click(function(event) {
event.preventDefault();
$.post('index.php', $("form").serialize(), function(data) {
// Whatever you want
console.log(data); // Open browser console to see your responsed data.
});
});