dp152153 2014-01-23 03:51
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PHP登录使用2d数组，并且使用$ .get或$ .post的jquery不刷新页面

I got here a working code that returns the corresponding user_id of the user when he logs in. But then, I need to use the $.get or $.post of jquery so that there will be no page refresh. I read posts about using $.get and $.post, but I doesn't seem to understand them.

Here is my working code. Please help me apply $.get or $.post to this. thanks!

<?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>
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2条回答 默认 最新

  • dsj83686 2014-01-23 04:10
    关注

    You need ajax for fulfill your requirement:

    1) You need to separate php login functionalities in a separate php file.
    2) you need to make an ajax call when clicking in login button. you need to add some javascript for your ajax call.
    3) add jquery in your html file

    Your php file( i have named it 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.';
    }

    in your html file:

    <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>

    You need to learn basic of ajax-php-jquery: http://blog.teamtreehouse.com/beginners-guide-to-ajax-development-with-php

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论
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