Question

我为登录系统编写了这个JavaScript代码。我在将其连接到PHP文件时遇到问题。我想从我的PHP脚本中显示一条消息并隐藏此表单。

    $('#submit').click(function()
    {
var email=$("#email").val();

var password=$("#password").val();
if(ck_email.test(email) && ck_password.test(password) )
{
$("#form").show().html("<h1>Thank you!</h1>");
}
return false;
});

PHP脚本

include ('database_connection.php');
if (isset($_POST['formsubmitted'])) {
if (empty($error)) //send to Database if there's no error '

    { // If everything's OK...

       // Make sure the email address is available:
       $query_verify_email = "SELECT * FROM members  WHERE Email ='$Email'";
       $result_verify_email = mysqli_query($dbc, $query_verify_email);
       if (!$result_verify_email) {            echo ' Database Error Occured ';
       }

       if (mysqli_num_rows($result_verify_email) == 0) { // IF no previous user is using this email .


        // Create a unique  activation code:
        $activation = md5(uniqid(rand(), true));


        $query_insert_user = "INSERT INTO `members` ( `Username`, `Email`, `Password`, `College`, `Activation`) VALUES ( '$name', '$Email', '$Password', '$College' '$activation')";


        $result_insert_user = mysqli_query($dbc, $query_insert_user);
        if (!$result_insert_user) {
        echo 'Query Failed ';
         }

        if (mysqli_affected_rows($dbc) == 1) { //If the Insert Query was successfull.


        // Send the email:
        $message = " To activate your account, please click on this link:\n\n";
        $message .= WEBSITE_URL . '/activate.php?email=' . urlencode($Email) . "&key=$activation";
        mail($Email, 'Registration Confirmation', $message, 'From: cr@gmail.com');

        // Flush the buffered output.


        // Finish the page:
        echo '<div class="success">Thank you for
registering! A confirmation email
has been sent to '.$Email.' Please click on the Activation Link to Activate your account </div>';


         } else { // If it did not run OK.
         echo '<div class="errormsgbox">You could not be registered due to a system
error. We apologize for any
inconvenience.</div>';
         }

        } else { // The email address is not available.
        echo '<div class="errormsgbox" >That email
address has already been registered.
</div>';
        }
    } else {//If the "error" array contains error msg , display them
echo '<div class="errormsgbox"> <ol>';
        foreach ($error as $key => $values) {

            echo '  <li>'.$values.'</li>';

}
echo '</ol></div>';
}
  mysqli_close($dbc);//Close the DB Connection
}

我知道我必须使用Ajax，但是，我无法获得正确的代码。

Answer 1

您必须向运行php脚本的网络服务器地址发送帖子或获取请求。你得到的答案就是PHP脚本的echo部分。

例如，使用JQuery和post：

    $.post(server+'index.php',{ email: $("#email").val(),formsubmitted:true, password: $("#password").val() }, function(data) {
        $("#form").show().html(data);
        });

在第一个post-parameter中，你会说出你的服务器所在的位置，以及你想要访问的php文件。通常是

   server = "http://127.0.0.1/"

作为服务器的localhost。在第二个参数中，您可以为PHP脚本提供后变量，PHP可以在$ POST ['email']中使用，例如可以使用。第三个参数是回调，当服务器的回答到达时，将执行回调。

当然，您仍然需要编辑上面发布的一些PHP以使其正常工作。例如，在使用之前从Post字段加载$ Email-Variable：

    $Email = $POST['email'];

如何使用ajax的submit函数运行PHP脚本？

1 个答案: