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Javascript Ajax没有从php获得响应_Javascript_Php_Jquery_Ajax - Fatal编程技术网

Javascript Ajax没有从php获得响应

Javascript Ajax没有从php获得响应,javascript,php,jquery,ajax,Javascript,Php,Jquery,Ajax,我真的不知道我的代码出了什么问题。php的响应并没有返回到我的ajax调用。这是我的密码。JQuery $(".theform").submit(function(e){ var formData = { "name": $('#name').val(), "email": $('#email').val(), "phone": $('#phone').val(),

我真的不知道我的代码出了什么问题。php的响应并没有返回到我的ajax调用。这是我的密码。JQuery

$(".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

当我检查请求时,我看到了来自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,我做错什么了?顺便说一句,它现在应该返回“failed”

请尝试在控制台中打印一次响应,并在ajax成功回调中尝试
=
而不是
=
。添加一个错误句柄如果您返回的是一个简单字符串“failed”,为什么要指定
数据类型:'json'
?谢谢……问题实际上是数据类型:'json'
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];