Php表单ajax“成功与失败”消息

Question

The form on my website is a simple contact form. I would like the form to show a "success & failed" message on the same page when the form is sent/failed without reloading the page. I understand that I should use Ajax to do this but I can't get it to work because my knowledge about it is very little.

Here is the code I'm working with.

Html (single page design):

    <script src="http://ajax.googleapis.com/ajax/libs/jquery/1.6.2/jquery.js"></script>

<form class="form" id="contactus" action="" method="post" accept-charset="UTF-8">




                        <label for="nametag">Namn<FONT COLOR="#FF0060">*</FONT></label>

                        <input name="name" type="text" id="name"  value="" />




                        <label for="emailtag">Email<FONT COLOR="#FF0060">*</FONT></label>

                        <input name="email" type="text" id="email"  value="" />


                        <label for="phonetag">Telefon</label>

                        <input name="phone" type="text" id="phone"  value="" />


                        <label for="messagetag">Meddelande<FONT COLOR="#FF0060">*</FONT></label></br>

                        <textarea name="message" id="message" style="width: 87%; height: 200px;"></textarea>






<label class="placeholder">&nbsp;</label>

                        <button class="submit" name="submit">Skicka</button>



                </form> 



<script>        
    $(function() {
            $('#contactus').submit(function (event) {
                event.preventDefault();
                event.returnValue = false;
                $.ajax({
                    type: 'POST',
                    url: 'php/mail.php',
                    data: $('#contactus').serialize(),
                    success: function(res) {alert(res);
                        if (res == 'successful') {
                            $('#status').html('Sent').slideDown();
                        }
                        else {
                            $('#status').html('Failed').slideDown();
                        } 
                    },
                    error: function () {
                        $('#status').html('Failed').slideDown();
                    }
                });
            });
        });
    </script>   

Php:

<?php
    $name = $_POST['name'];
    $email = $_POST['email'];
    $phone = $_POST['phone'];
    $message = $_POST['message'];
    $recipient = "info@mydomain.com";
    $subject = "Webbkontakt";
    $formcontent = "Från: $name <br/> Email: $email <br/> Telefon: $phone <br/> Meddelande: $message";

    $headers = "From: " ."CAMAXON<info@mydomain.com>" . "\r\n";
    $headers .= "Reply-To: ". "no-reply@mydomain.com" . "\r\n";
    $headers .= "MIME-Version: 1.0\r\n";
    $headers .= "Content-Type: text/html; charset=utf-8\r\n";

    if(mail($recipient, $subject, $formcontent, $headers))
    {
        echo "successful";
    }
    else
    {
        echo "error";
    }
?>

Answer 1

Your Ajax call is not working properly. Try this

$(function() {
            $('#contactus').submit(function (event) {
                event.preventDefault();
                event.returnValue = false;
                $.ajax({
                    type: 'POST',
                    url: 'php/mail.php',
                    data: $('#contactus').serialize(),
                    success: function(res) {
                        if (res == 'successful') {
                            $('#status').html('Sent').slideDown();
                        }
                        else {
                            $('#status').html('Failed').slideDown();
                        } 
                    },
                    error: function () {
                        $('#status').html('Failed').slideDown();
                    }
                });
            });
        });

Also as you can see i have used $('#contactus').serialize() this way you dont need to pass the form elements one by one instead serialize() will pass the whole form elements to your php page

Than in your php file echo successful if everything went well else echo error if the response is an error than show the error div

<?php
    $name = $_POST['name'];
    $email = $_POST['email'];
    $phone = $_POST['phone'];
    $message = $_POST['message'];
    $recipient = "info@mydomain.com";
    $subject = "Webbkontakt";
    $formcontent = "Från: $name <br/> Email: $email <br/> Telefon: $phone <br/> Meddelande: $message";

    $headers = "From: " ."CAMAXON<info@mydomain.com>" . "\r\n";
    $headers .= "Reply-To: ". "no-reply@mydomain.com" . "\r\n";
    $headers .= "MIME-Version: 1.0\r\n";
    $headers .= "Content-Type: text/html; charset=ISO-8859-1\r\n";

    if(mail($recipient, $subject, $formcontent, $headers))
    {
        echo "successful";
    }
    else
    {
        echo "error";
    }
?>

Answer 2

Change your PHP script like this:

<?php
if( isset( $_POST['submit'] )){ //checking if form was submitted
$name = $_POST['name'];
$email = $_POST['email'];
$phone = $_POST['phone'];
$message = $_POST['message'];
$formcontent="Meddelande: \n\n $message";
$recipient = "info@mydomain.com";
$subject = "Webbkontakt";
$mailheader = "Från: $name \n Email: $email \n Telefon: $phone \r\n";

$mailsent = mail($recipient, $subject, $formcontent, $mailheader);

if($mailsent) echo "Success"; //success if mail was sent
else echo "Ett fel uppstod!";
}
?>

Answer 3

在您的mail()函数下方，只需执行echo "successful";

Answer 4

2020 Edit

In REST API response should be always accompanied by the correct HTTP status code, with 200+ telling client that request ended up correctly processed or was otherwise good, 400+ telling client there was an error in request, and 500+ that there was a problem with the server itself. Do not use statuses inside responses, it is unnecessary duplication of the existing feature:

http_response_code(200);
echo json_encode(['message' => 'Email was sent']);

exit;

Then you can handle request and response with jQuery(assuming that you still use jQuery):

$.ajax({
  url: url,
  data: data,
  dataType: 'json'
})
  .then(function (data, textStatus, jqXHR) {
    // Your 200+ responses will land here
  })
  .fail(function (jqXHR, textStatus, errorThrown) {
    // Your 400+ responses will be caught by this handler
  });
;

If you need specific status, you can get it from jqXHR parameter using jqXHR.status field.

Original answer

You can use dataType: 'json' in your ajax call.

Then you'll be able to pass status code as response key:

// form response array, consider it was success
$response = array( 'success'=> 'ok', 'message' => 'Email was sent');

echo json_encode($response);

In js then you can check data.success === 'ok' to know what's your status.

Answer 5

In your php script you could try this

if(mail($recipient, $subject, $formcontent, $mailheaders))
{
  echo("Mail Sent Successfully"); // or echo(successful) in your case
}else{
  echo("Mail Not Sent"); // or die("Ett fel uppstod!");

}