gpt4 book ai didi

php - 弹出表单ajax请求不起作用

转载 作者:可可西里 更新时间:2023-11-01 00:08:37 24 4
gpt4 key购买 nike

我想在我的站点中添加弹出式联系表单,我从互联网上举了一个例子,并对其发送邮件进行了一些更改,并将其用于将数据插入数据库。所以这是我的第一个文件 index.php

<!doctype html>
<html lang="en">
<head>

<title>A Slick Ajax Contact Form with jQuery and PHP</title>
<meta http-equiv="Content-Type" content="text/html;charset=utf-8">
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.5.1/jquery.min.js"></script>

<script type="text/javascript">

var messageDelay = 2000; // How long to display status messages (in milliseconds)

// Init the form once the document is ready
$(init);

// Initialize the form
function init() {

// Hide the form initially.
// Make submitForm() the form's submit handler.
// Position the form so it sits in the centre of the browser window.
$('#contactForm').hide().submit(submitForm).addClass('positioned');

// When the "Send us an email" link is clicked:
// 1. Fade the content out
// 2. Display the form
// 3. Move focus to the first field
// 4. Prevent the link being followed

$('a[href="#contactForm"]').click(function() {
$('#content').fadeTo('slow', .2);
$('#contactForm').fadeIn('slow', function() {
$('#senderName').focus();
})

return false;
});

// When the "Cancel" button is clicked, close the form
$('#cancel').click(function() {
$('#contactForm').fadeOut();
$('#content').fadeTo('slow', 1);
});

// When the "Escape" key is pressed, close the form
$('#contactForm').keydown(function(event) {
if (event.which == 27) {
$('#contactForm').fadeOut();
$('#content').fadeTo('slow', 1);
}
});

}

// Submit the form via Ajax
function submitForm() {

var contactForm = $(this);

// Are all the fields filled in?
if (!$('#senderName').val() || !$('#senderEmail').val() || !$('#ContactNo').val() || !$('#message').val()) {

// No; display a warning message and return to the form
$('#incompleteMessage').fadeIn().delay(messageDelay).fadeOut();
contactForm.fadeOut().delay(messageDelay).fadeIn();

} else {

// Yes; submit the form to the PHP script via Ajax
$('#sendingMessage').fadeIn();
contactForm.fadeOut();

$.ajax({
url: contactForm.attr('action') + "?ajax=true",
type: contactForm.attr('method'),
data: contactForm.serialize(),
success: submitFinished
});

}

// Prevent the default form submission occurring
return false;
}

// Handle the Ajax response
function submitFinished(response) {
response = $.trim(response);
$('#sendingMessage').fadeOut();

if (response == "success") {

// Form submitted successfully:
// 1. Display the success message
// 2. Clear the form fields
// 3. Fade the content back in

$('#successMessage').fadeIn().delay(messageDelay).fadeOut();
$('#senderName').val("");
$('#senderEmail').val("");
$('#ContactNo').val("");
$('#message').val("");

$('#content').delay(messageDelay + 500).fadeTo('slow', 1);

} else {

// Form submission failed: Display the failure message,
// then redisplay the form
$('#failureMessage').fadeIn().delay(messageDelay).fadeOut();
$('#contactForm').delay(messageDelay + 500).fadeIn();

}

}

</script>

</head>
<body>

<div id="content">
<p style="padding-bottom: 50px; font-weight: bold; text-align: center;"><a href="#contactForm">~ Send us an email ~</a></p>
</div>

<form id="contactForm" action="processForm.php" method="post">
<h2>Send us an email...</h2>
<ul>
<li>
<label for="senderName">Full Name</label>
<input type="text" name="senderName" id="senderName" placeholder="Please type your name" required="required" maxlength="40" />
</li>
<li>
<label for="senderEmail">Email</label>
<input type="email" name="senderEmail" id="senderEmail" placeholder="Please type your email address" required="required" maxlength="50" />
</li>
<li>
<label for="ContactNo">Contact No</label>
<input type="text" name="ContactNo" id="ContactNo" placeholder="Please type your contact no" required="required" maxlength="50" />
</li>
<li>
<label for="message" style="padding-top: .5em;">Your Message</label>
<textarea name="message" id="message" placeholder="Please type your message" required="required" cols="80" rows="10" maxlength="10000"></textarea>
</li>
</ul>
<div id="formButtons">
<input type="submit" id="sendMessage" name="sendMessage" value="Send Email" />
<input type="button" id="cancel" name="cancel" value="Cancel" />
</div>
</form>

<div id="sendingMessage" class="statusMessage">
<p>Sending your message. Please wait...</p>
</div>

<div id="successMessage" class="statusMessage">
<p>Thanks for sending your message! We'll get back to you shortly.</p>
</div>

<div id="failureMessage" class="statusMessage">
<p>There was a problem sending your message. Please try again.</p>
</div>

<div id="incompleteMessage" class="statusMessage">
<p>Please complete all the fields in the form before sending.</p>
</div>

</body>
</html>

还有一个是processForm.php。

<?php

// Read the form values
$success = false;
$senderName = isset( $_POST['senderName'] ) ? preg_replace( "/[^\.\-\' a-zA-Z0-9]/", "", $_POST['senderName'] ) : "";
$senderEmail = isset( $_POST['senderEmail'] ) ? preg_replace( "/[^\.\-\_\@a-zA-Z0-9]/", "", $_POST['senderEmail'] ) : "";
$ContactNo = isset( $_POST['ContactNo'] ) ? preg_replace( "/[^\ 0-9]/", "", $_POST['ContactNo'] ) : "";
$message = $_POST['message'];

// If all values exist, send the email
if ( $senderName && $senderEmail && $contactNo && $message ) {

include("config.php");
$success = mysql_query("INSERT INTO `inquiry` (`name`, `contact`, `email`, `query`) VALUES ('$senderName', '$ContactNo', '$senderEmail', '$message')");

}

// Return an appropriate response to the browser
if ( isset($_GET["ajax"]) ) {

echo $success ? "success" : "error";

} else { ?>

<html>
<head>
<title>Thanks!</title>
</head>
<body>

<?php if ( $success ) echo "<p>Thanks for sending your message! We'll get back to you shortly.</p>" ?>
<?php if ( !$success ) echo "<p>There was a problem sending your message. Please try again.</p>" ?>
<p>Click your browser's Back button to return to the page.</p>

</body>
</html>

<?php } ?>

问题是当我提交正确时显示消息“发送您的消息时出现问题。请重试。”请帮助我。

最佳答案

我认为你应该改变这一行。

if ( $senderName && $senderEmail && $contactNo && $message ) {

if ( $senderName && $senderEmail && $ContactNo && $message ) {

关于php - 弹出表单ajax请求不起作用,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/33318817/

24 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com