PHP post请求数据检索使用Jquery AJAX - PHP post request data retrieval using Jquery AJAX

PHP post request data retrieval using Jquery AJAX

我最近一直在尝试使用Jquery的ajax函数从php获取信息发送请求工作正常，但接收信息不。谢谢你的帮助。

Jquery:

$.ajax({                    
  url: 'sendEmail.php',     
  type: 'post', // performing a POST request
  data: values,
  dataType: 'json',                   
  success: function(result)         
  {
    console.log("check");//code does not come here
    if(result=="Success"){
        alert("Your message has been sent. I will get in touch with you soon");
    }
    else{
        alert("Umm.. The meesage was not sent :(. You can still contact me on myemail@gmail.com");
    }
  } 
});

php:

<?php
   $recipient = 'myemail@gmail.com';
   $subject = "Portfolio Message";
   $fromName = stripslashes($_POST['Name']);
   $fromEmail= stripcslashes($_POST['Email']);
   $msg = "Message from: $fromMessage'nEmail: $fromEmail'n'n".stripslashes($_POST['Message']);

   if (mail($recipient, $subject, $msg)){
       echo "Success";
   }
   else {
       echo "Fail";
   }
?>

问题:

响应类型与ajax期望的数据类型不匹配

你有两个选择:

将数据类型更改为文本
发送一个正确的json响应

两者中的任何一个都将解决您的问题发送json的示例是在json_encode

中编码响应

dataType: 'json',

需要PHP的json回复，在PHP代码中添加header。

<?php
   header('Content-Type: application/json');
   $recipient = 'myemail@gmail.com';
   $subject = "Portfolio Message";
   $fromName = stripslashes($_POST['Name']);
   $fromEmail= stripcslashes($_POST['Email']);
   $msg = "Message from: $fromMessage'nEmail: $fromEmail'n'n".stripslashes($_POST['Message']);
   if (mail($recipient, $subject, $msg)){
       echo json_encode("Success");
   }
   else {
       echo json_encode("Fail");
   }
?>