php后端对ajax请求的响应应该如何？

Posted 2023-05-08

【中文标题】php后端对ajax请求的响应应该如何？

【英文标题】：How should be the response to an ajax request in php backend?

【发布时间】：2020-12-20 11:19:21

【问题描述】：

我试图在客户端取得成功，但我做不到。如果我把

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

ajax 请求中的这段代码捕获了一些东西，但它不应该出错。

我尝试了很多东西，但找不到解决方案。

这是我在客户端的 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>

这里是后端；

<?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;
    

?>

【参考方案1】：

我会尝试以这种方式执行 javascript，如示例中所示：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>

【讨论】：

我使用了 statusCode: 200: function()..... 并且成功了。非常感谢。