mankutila mankutila - 6 months ago 20
jQuery Question

Sending e-mail form using PHP and AJAX

I have contact form on my site. It sends message to email. I try to do it without page reload using AJAX, but it seems that AJAX doesn't work: messages are sent but the page still redirecting to

call-form.php
. What is incorrect in my code? (jQuery is included)

HTML

<form name="freeCall" action="<?php bloginfo(template_url); ?>/mail/call-form.php" method="post" class="popover-form" id="free-call-form">
<label for="name1">Name</label><span class="pull-right close">&times;</span><input placeholder="Name" name="call-name" type="text" id="name1" >
<label for="phone">Phonenumber</label><input name="phone" type="text" value="" placeholder="+375" id="phone" >
<input type="submit" value="Call me back" >
</form>


PHP - call-form.php

<?
if((isset($_POST['call-name']))&&(isset($_POST['phone'])&&$_POST['phone']!="")){
$to = 'test@gmail.com';
$subject = 'Callback';
$message = '
<html>
<head>
<title>Call me back</title>
</head>
<body>
<p><b>Name:</b> '.$_POST['call-name'].'</p>
<p><b>Phonenum:</b> '.$_POST['phone'].'</p>
</body>
</html>';
$headers = "Content-type: text/html; charset=utf-8 \r\n";
$headers .= "From: Site <info@mail.com>\r\n";
mail($to, $subject, $message, $headers);
}
?>


JS

$(function () {
$("#free-call-form").submit(function () {
var form_data = $(this).serialize();
$.ajax({
type: "POST",
url: "call-form.php",
data: form_data,
success: function () {
alert("It's OK!");
}
});
});
});

Answer

Ok, first when you make an AJAX call, you must have a way to know if your PHP returns you something (useful for debugging).

Then, when submitting a form with AJAX, the tag action="" is not needed.

Finally, to prevent a form from being sent when making an AJAX call, add e.preventDefault() with the event called e here, like in my example.

I have improved your code to be more realistic about the latest standards.

HTML :

<form name="freeCall" method="post" class="popover-form" id="free-call-form">  
  <label for="name1">Name</label><span class="pull-right close">&times;</span><input placeholder="Name" name="call-name" type="text" id="name1" >  
  <label for="phone">Phonenumber</label><input name="phone" type="text" value="" placeholder="+375" id="phone" >    
  <input type="submit" value="Call me back" >       

JS :

$(function () {    
  $("#free-call-form").submit(function (e) {
    e.preventDefault();
    var form_data = $(this).serialize(); 
    $.ajax({
      type: "POST", 
      url: "call-form.php",
      dataType: "json", // Add datatype
      data: form_data
    }).done(function (data) {
        console.log(data);
        alert("It's OK!");
    }).fail(function (data) {
        console.log(data);
    });
  }); 
});

And PHP :

if((isset($_POST['call-name']))&&(isset($_POST['phone'])&&$_POST['phone']!="")){ 
  $to = 'test@gmail.com';
  $subject = 'Callback';
  $message = '
        <html>
            <head>
                <title>Call me back</title>
            </head>
            <body>
                <p><b>Name:</b> '.$_POST['call-name'].'</p>
                <p><b>Phonenum:</b> '.$_POST['phone'].'</p>                        
            </body>
        </html>'; 
  $headers  = "Content-type: text/html; charset=utf-8 \r\n"; 
$headers .= "From: Site <info@mail.com>\r\n"; 
mail($to, $subject, $message, $headers);

  echo json_encode(array('status' => 'success'));
} else {
  echo json_encode(array('status' => 'error'));
}

With echo json_encode, you know what is the return of your AJAX call. It is better