在php后台如何响应ajax请求？

Question

我试图在客户端获得成功，但我做不到。如果我把

error:function(data){
                console.log(data);
                console.log('error');
            }

ajax中的这段代码请求this捕获了一些东西，但它不应该出错。

我试了那么多东西，但还是找不到解决方案。

这里是我在客户端的ajax请求；

<script>
$(document).on("submit", "#request-form", function(event){ //request-form id li form post edildiğinde
    event.preventDefault();
    var serialized = $(this).serialize();
    if(serialized.indexOf('=&') > -1 || serialized.substr(serialized.length - 1) == '='){ //formda boş yer var ise
        alert("Fill in all fields");
    }else{
        $.ajax({
            url: "http://127.0.0.1/rent_website/mail-sender/mail.php", //"https://stanstedcab.co.uk/project/mail-sender/mail.php", 
            type: "POST",             
            data: serialized,
            dataType: "json",
            function(data, status) {
                console.log('function works');
                if (data.success) {
                    console.log(data);
                    console.log('Başarılı');
                } else {
                    console.log('else', data)
                }
            },
        });
    }
});
</script>

和这里的后端；

<?php
$response = array();

if ($_POST){
    if(isset($_POST["your-pickup"]) && isset($_POST["your-drop"]) && isset($_POST["Vehicle"]) && 
            isset($_POST["meeting-time"]) && isset($_POST["your-name"]) && isset($_POST["your-phone"]) && 
            isset($_POST["your-email"]) && isset($_POST["your-message"])) {

        //here some mail settings

        if($mail->Send()){
            $message = "Email sent";
            $response["success"] = true;
            $response["message"] = $message;
            echo json_encode($response);
            return $response;
        } else {
            $response = array('result' => 'Email couldn\'t sent', 'success' => false);
            echo json_encode($response);
            return $response;
        }
    }else{
        $response = array('result' => 'Fill all fields.', 'success' => false);
        echo json_encode($response);
        return $response;
    }
}
?>

Answer 1

我试着用下面示例中所示的方式编写Javascript：https://api.jquery.com/jquery.ajax/

<script>
$(document).on("submit", "#request-form", function(event){ //request-form id li form post edildiğinde
    event.preventDefault();
    var serialized = $(this).serialize();
    if(serialized.indexOf('=&') > -1 || serialized.substr(serialized.length - 1) == '='){ //formda boş yer var ise
        alert("Fill in all fields");
    }else{
        $.ajax({
            url: "http://127.0.0.1/rent_website/mail-sender/mail.php", //"https://stanstedcab.co.uk/project/mail-sender/mail.php", 
            type: "POST",             
            data: serialized,
            dataType: "json"
       }).done(function(data) {
                console.log('function works');
                if (data.success) {
                    console.log(data);
                    console.log('Başarılı');
                } else {
                    console.log('else', data)
                }
        });
    }
});
</script>