0

我有几个使用相同的通用 AJAX 代码的联系表格：

$(document).ready(function() {
    $('#mainMail').submit( function() {
        $.ajax({
            type: "POST",

            url : "<?php echo home_url(); ?>/wp-content/themes/retlehs-roots-f45f4f5/assets/bin/process.php",
            data: $(this).serialize(),
            success: function() {
                    $('#enquiryBox #submitSuccess').removeClass('hidden');
                    setTimeout(function() { $("#enquiryBox #submitSuccess").addClass('hidden'); }, 15000);       
                },
            error: function() {
                $('#enquiryBox #submitFail').removeClass('hidden');
                setTimeout(function() { $("#enquiryBox #submitFail").addClass('hidden'); }, 15000);
            }
            });
        return false;
    });
});

简而言之，当它成功提交表单并向我发送电子邮件（通过 process.php）时，#submitSuccessdiv 会按应有的方式显示出来。问题是，当它失败时，div也会出现（#submitFaildiv 保持隐藏状态）。

我猜这是因为只有在没有成功调用 process.php 时才会发生 ajax 发布失败？有没有办法调整它，以便如果电子邮件不是通过 process.php 发送给我，用户会收到错误消息？

这是我的 process.php 代码：

<?php 

$robotest = $_POST['robotest']; //just testin' for robots

$recipient = "info@mydomain.com"; //recipient 
$email = ($_POST['email']); //senders e-mail adress 

if((filter_var($email, FILTER_VALIDATE_EMAIL)) && ($robotest == "")) { 

$Name = ($_POST['name']); //senders name 

$mail_body  = "Hello, \r\n \r\n";
$mail_body .= "You have received a new booking with the following details: \r\n\r\n";
$mail_body .= "{$_POST['comments']}; \r\n \r\n";
$mail_body .= "Please respond to the customer within 30 minutes on the following phone number: {$_POST['phone']} \r\n";
$mail_body .= "Warm regards, \r\n";
$mail_body .= "Robot. \r\n";



$subject = "Free Consult Inquiry"; //subject 
$header = "From: ". $Name . " <" . $email . ">\r\n"; //optional headerfields 

mail($recipient, $subject, $mail_body, $header); //mail command :) 

} else {
  print "You've entered an invalid email address!";
 }
?>

（这print "you've entered an invalid email address!"是我对错误消息的尝试，但这也不起作用......）

4

3 回答 3

1

将此行也添加$("#enquiryBox #submitSuccess").addClass('hidden');到错误块中，

  $.ajax({
        type: "POST",

        url : "<?php echo home_url(); ?>/wp-content/themes/retlehs-roots-f45f4f5/assets/bin/process.php",
        data: $(this).serialize(),
        success: function() {
                $('#enquiryBox #submitSuccess').removeClass('hidden');
                setTimeout(function() { $("#enquiryBox #submitSuccess").addClass('hidden'); }, 15000);       
            },
        error: function() {
            $('#enquiryBox #submitFail').removeClass('hidden'); //why do you remove class 'hidden' and add it later too ??
            $("#enquiryBox #submitSuccess").addClass('hidden'); //added line here
            setTimeout(function() { $("#enquiryBox #submitFail").addClass('hidden'); }, 15000);
        }
        });
于 2012-11-12T22:53:46.337 回答
1

您可以通过返回 HTTP 错误代码来指示错误，例如

} else {
  header("HTTP/1.1 500 Internal error");
  print "You've entered an invalid email address!";
 }
于 2012-11-12T23:23:56.267 回答
0

如果您阅读 上的手册条目jQuery.ajax()，您会发现错误事件发生在发生协议或网络错误时。如果您的脚本 return或类似的东西false，null它仍然被认为是有效的响应，因此您的问题。

在这种情况下，我会编写它，以便所有调用最终都调用一个方法，如果调用错误事件，它将调用相同的方法并显示重建的错误消息。

于 2012-11-12T22:56:34.473 回答