Question

我真的不知道我的代码在这里有什么问题。来自php的回复没有回到我的ajax调用。这是我的代码。 JQuery的

$(".theform").submit(function(e){
            var formData = {
                "name": $('#name').val(),
                "email": $('#email').val(),
                "phone": $('#phone').val(),
                "message": $('#message').val()
            }

            var sendMail = $.ajax({
                type: 'POST',
                url: 'submit.php',
                data: formData,
                dataType: 'json',
                encode: true
            });

            sendMail.done(function(output){
                if(output == "failed"){
                        console.log("failed");
                )
                        }
                });

            e.preventDefault();
        });

PHP

if($_POST){

$adminEmail = "";
$data = [];
$to = $adminEmail;
$name = $_POST['name'];
$email = $_POST['email'];
$number = $_POST['phone'];
$headers = "From: $email". "\r\n" .
"Reply-To: $email" . "\r\n" .
'X-Mailer: PHP/' . phpversion();

$message = $_POST['message'];

$sendMail = mail($to, "Mail from $name", $message, $headers);


if($sendMail){
    $data = ["success"];
} else {
    $data = ["failed"];
   }
 }
print $data[0];

当我检查请求时，我看到来自php的响应，但我似乎无法在ajax调用中将其打印到屏幕或控制台。拜托，我做错了什么？顺便说一句，它应该返回＆＃34;失败＆＃34;现在

Ajax没有得到php的响应

0 个答案: