Question

我是php的新手，似乎无法理解我的代码中的问题：

<?php
    require('mysql_connect.php');
    // If the values are posted, insert them into the database.
    if (isset($_POST['username']) && isset($_POST['password'])&& isset($_POST['phone'])){
        $username = $_POST['username'];
        $phone = $_POST['phone'];
        $email = $_POST['email'];
        $password = $_POST['password'];

        $query = "INSERT INTO `user` (username, password, email,phone) VALUES ('$username', '$password', '$email', '$phone')";
        $result = mysql_query($query);
        if($result){
            $msg = "User Created Successfully.";
            }

        echo "
        <script >
        function changeIT() {
        $( '#phone,#reg' ).css('display','inline');
        }
        $(document).ready(changeIT);

        </script>";
        header('Location: http://atestat-cpi.comeze.com/index.html');
     }
?>

根据我对php的理解，这应该执行jquery来显示所需的元素，然后显示带有显示元素的index.html。问题是，当页面被加入时，它什么都不显示，只是空白窗口，所以我认为jQuery没有执行。请帮助。

Answer 1

header('Location: /')进行重定向，因此不会发生任何事情。您必须重新排序代码和文件。我将通过一个示例向您展示一种继续进行的方式。

首先，让我们准备你的索引文件。我将直接使用ajax而不是典型的表单提交。 这是一个例子：

<!DOCTYPE html>
<html>
<head>
    <!-- load jQuery -->
    <script type="text/javascript" src="//ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

    <!-- call to your own js -->
    <script type="text/javascript" src="register_user.js"></script>
</head>
<body>
    <!-- Instead of create a form you can do so -->
    <div id="register">
        <label><input id="user" type="text" value="" /></label>
        <label><input id="pass" type="password" value="" /></label>
        <label><input id="phone" type="phone" value="" /></label>
        <label><input id="email" type="email" value="" /></label>
        <input id="register_user" type="button" value="register!" />
    </div>
</body>
</html>

现在，让我们设置您的javascript文件，我称之为register_user.js：

// on dom ready
$(function(){
    $('#register_user').on('click', function(){

        // disable button, prevent multiple calls
        var $this = $(this);
        $this.prop('disabled', true);

        $.ajax({
            url: 'register.php',
            type: 'post',
            data: {
                username: $('#user').val(),
                password: $('#pass').val(),
                phone: $('#phone').val(),
                email: $('#email').val()
            },
            success: function(data){
                var response = data.split('|');
                if( response[0] == '1' ){

                    // this is part of your code, it should be in your index file
                    $('#phone, #reg').css('display', 'inline');

                    // restore the button (if you want) or whatever you want to do .)
                    $this.prop('disabled', false);
                }
                else{
                    alert(response[1]);
                }
            }
        });
    });
});

现在让我们制作你的php用户注册脚本。这应该是一个独立的脚本，我称之为register.php：

<?php
$responses = array(
    '-1|One or more required fields were not filled',
    '0|There was an error creating the user. Try again.',
    '1|Your user account was created successfully!'
);

if( isset($_POST['username'], $_POST['password'], $_POST['phone'], $_POST['email']) ){
    require_once 'mysql_connect.php';

    $username = $_POST['username'];
    $password = $_POST['password'];
    $phone = $_POST['phone'];
    $email = $_POST['email'];

    // You MUST use mysqli, but now let's use your current code
    $query = "INSERT INTO `user` (username, password, email,phone) VALUES ('$username', '$password', '$email', '$phone')";
    $result = mysql_query($query);

    // If the user was created successfully, say ok.
    // The usually is to say yes or no (1 or 0) and manage the message with JS in the view.
    if( $result ) echo responses[2];
    else echo responses[1];
    exit();
}
echo responses[0];
exit();
?>

PS：我没有测试这段代码，所以它可能有失败或需要修复的东西。不记得它只是一个例子。

Answer 2

试试这样..

<?php
  //php code..
?>
<script>
  <!--your script-->
</script>
<?php
  //php code..
?>

如果您想要jQuery工作，则还应包括jquery library。

Answer 3

键入header('Location: http://atestat-cpi.comeze.com/index.html');后，您将被重定向到该uri，因此您的回音无效。此外，您似乎不包括JQuery，您可以使用CDN https://jquery.com/download/#using-jquery-with-a-cdn包含它。此代码中也存在SQL注入的主要问题。您应该使用mysql_real_escape_string(input)方法转义用户输入。

php注册脚本与jquery不能正常工作

3 个答案: