Question

我正在尝试让PHP在提交表单时将呈现的HTML注入页面。表单数据用AJAX拦截，然后发送给PHP。

这是HTML表单：

<span id="message-sent">
     <form role="form" id="footer-form" name="contactform" method="post" action="email.php">
                                <div class="form-group has-feedback">
                                    <label class="sr-only" for="name">Name</label>
                                    <input type="text" class="form-control" id="name" placeholder="Name" name="name" required>
                                    <i class="fa fa-user form-control-feedback"></i>
                                </div>
                                <div class="form-group has-feedback">
                                    <label class="sr-only" for="email">Email address</label>
                                    <input type="email" class="form-control" id="email" placeholder="Enter email" name="email" required>
                                    <i class="fa fa-envelope form-control-feedback"></i>
                                </div>
                                <div class="form-group has-feedback">
                                    <label class="sr-only" for="message">Message</label>
                                    <textarea class="form-control" rows="8" id="message" placeholder="Message" name="message" required></textarea>
                                    <i class="fa fa-pencil form-control-feedback"></i>
                                </div>
                                <input type="submit" value="Send" class="btn btn-default">
                            </form>
    </span>

这是PHP：

<?php

// Added input sanitizing to prevent injection

// Only process POST reqeusts.
if ($_SERVER["REQUEST_METHOD"] == "POST") {
    // Get the form fields and remove whitespace.
    $name = strip_tags(trim($_POST["name"]));
            $name = str_replace(array("\r","\n"),array(" "," "),$name);
    $email = filter_var(trim($_POST["email"]), FILTER_SANITIZE_EMAIL);
    $message = trim($_POST["message"]);

    // Check that data was sent to the mailer.
    if ( empty($name) OR empty($message) OR !filter_var($email, FILTER_VALIDATE_EMAIL)) {
        // Set a 400 (bad request) response code and exit.
        echo "There was a problem with your submission. Please complete the form and try again.";
        exit;
    }

    // Set the recipient email address.
    $recipient = "example@example.com";

    // Set the email subject.
    $subject = "Website Message from $name";

    // Build the email content.
    $email_content = "Name: $name\n";
    $email_content .= "Email: $email\n\n";
    $email_content .= "Message:\n$message\n";

    // Build the email headers.
    $email_headers = "From: $name <$email>";

    // Send the email.
    if (mail($recipient, $subject, $email_content, $email_headers)) {
        // Set a 200 (okay) response code.
        echo 'Thank you for sending us a message!  We will get back to you ASAP!';
    } else {
        // Set a 500 (internal server error) response code.
        echo "Something went wrong and we couldn't send your message.";
    }

} else {
    // Not a POST request, set a 403 (forbidden) response code.
    echo "There was a problem with your submission, please try again.";
}

＆GT;

我希望服务器到页面的响应是HTML格式，然后通过AJAX插入原始页面。 AJAX填补了html代码中的范围。有任何想法吗？谢谢！

Answer 1

你可以这样做：

$.ajax({
    type: "POST",
    url: "ajax.php",
    data: {},
    }).done(function( htmlFromAjax ) {
        $("#ErrorMessageDiv").html(htmlFromAjax);
    }).fail(function(jqXHR, textStatus) {

    });

使用AJAX通过PHP呈现HTML

1 个答案: