AJAX/PHP form not sending information_问答_开发者

开发者 https://www.devze.com 2023-04-11 13:39 出处：网络

Hey guys i have a form that\'s supposed to send a message without reloading the page. I\'ve used this tutorial here: http://www.elated.com/articles/slick-ajax-contact-form-jquery-php/ and adapted it t

相关专题：jquery php

Hey guys i have a form that's supposed to send a message without reloading the page. I've used this tutorial here: http://www.elated.com/articles/slick-ajax-contact-form-jquery-php/ and adapted it to suit my needs. here is the php:

<?php
session_start();
// Define some constants
define( "RECIPIENT_NAME", "John Smith" );
define( "RECIPIENT_EMAIL", "myemail@email.com" );
define( "EMAIL_SUBJECT", "Visitor Message" );

// Read the form values and define variables
$success = false;
$senderName = isset( $_POST['name'] ) ? preg_replace( "/[^\.\-\' a-zA-Z0-9]/", "",    $_POST['name'] ) : "";
$senderEmail = isset( $_POST['email'] ) ? preg_replace( "/[^\.\-\_\@a-zA-Z0-9]/", "", $_POST['email'] ) : "";
$message = isset( $_POST['message'] ) ? preg_replace( "/(From:|To:|BCC:|CC:|Subject:|Content-Type:)/", "", $_POST['message'] ) : "";
$honeypot = $_POST['url'];
$code = $_POST['code'];

//if the honeypot is filled out, dont send the form
if (!empty($honeypot)){
    $success = false; 
}
// If all values exist and the code matches, send the email 
else if ( $senderName && $senderEmail && $message && $code == $_SESSION['random_code'] ) {
$recipient = RECIPIENT_NAME . " <" . RECIPIENT_EMAIL . ">";
$headers = "From: " . $senderName . " <" . $senderEmail . ">";
$success = mail( $recipient, EMAIL_SUBJECT, $message, $headers );
}



// Return an appropriate response to the browser
if ( isset($_POST["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
}
?>

and here is the javascript:

    function submitForm() {
 var contactForm = $(this);

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

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


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

function submitFinished( response ) {
 response = $.trim( response );
 $('#sendingMessage').fadeOut();

 if ( response == "success" ) {

// Form submitted successfully:
// 1. Display the success message
// 2. Clear the form fields

$('#succes开发者_运维问答sMessage').fadeIn().delay(5000).fadeOut();
$('#name').val( "" );
$('#email').val( "" );
$('#message').val( "" );

 } else {

// Form submission failed: Display the failure message,
$('#failureMessage').fadeIn().delay(5000).fadeOut();
}
}

that function is run when the user clicks the send button and after it's checked that all the fields have been filled out. I think it's tripping up down the end of the php where it says if(isset($_POST['ajax'])) because i keep getting redirected to the html fallback back, as if I dont have javascript enabled (which i do by the way). you can see a working example here: mattsandersdesign.com/test/contact2.html thanks in advance!

You have to return false even if all the validation checks pass. Otherwise the form will just submit.
Also are you even calling submitForm() ever?

ok i finally figured it out. firstly I was calling return false at the wrong time. I needed to run it outside the submitForm function. and secondly i had a couple of variables in the wrong places