使用JSON从PHP获取数据 [英] Getting data from a PHP with JSON

问题描述

我收到此错误：



可选（< meta charset = \UTF-8 \> {\电子邮件\ ：\ Error\ \ Password\：\ Error\} \r\\\
\r\\\
\r\\\
）<无线电通信/>
错误域= NSCocoaErrorDomain代码= 3840JSON文本没有以数组或对象开头，并且选项允许未设置片段。 UserInfo = {NSDebugDescription = JSON文本没有以数组或对象开头，并且选项允许未设置片段。}



我的服务器发送的回复类似于< ; meta charset =UTF-8> {电子邮件：错误，密码：错误}。

绝对不是有效的JSON，因为它开始于`<`





i不知道为什么它不起作用因为我有日志正确使用HTML代码从头开始



我的PHP代码

I'm getting this error:

Optional("<meta charset=\"UTF-8\">{\"Email\":\"Error\",\"Password\":\"Error\"}\r\n\r\n\r\n")
Error Domain=NSCocoaErrorDomain Code=3840 "JSON text did not start with array or object and option to allow fragments not set." UserInfo={NSDebugDescription=JSON text did not start with array or object and option to allow fragments not set.}

my server is sending a response like <meta charset="UTF-8">{"Email":"Error","Password":"Error"}.
It definitely is not a valid JSON, as it starts with `<`


i don't know why it's not working because i had log in correctly with HTML code from scratch

my php code

<?php  
  
// Create connection  
$con=mysqli_connect("cantshowthis","hehe","sorry",":c");  
  
// Check connection  
if (mysqli_connect_errno())  
{  
  echo "Failed to connect to MySQL: " . mysqli_connect_error();  
}  
  
// This SQL statement selects ALL from the table 'Registro'  
$sql = "SELECT * FROM Registro";  
  
// Check if there are results  
if ($result = mysqli_query($con, $sql))  
{  
  // If so, then create a results array and a temporary one  
  // to hold the data  
  $resultArray = array();  
  $tempArray = array();  
  
  // Loop through each row in the result set  
  while($row = $result->fetch_object())  
  {  
  // Add each row into our results array  
  $tempArray = $row;  
  array_push($resultArray, $tempArray);  
  }  
  
  // Finally, encode the array to JSON and output the results  
  echo json_encode($resultArray);  
}  
  
// Close connections  
mysqli_close($con);  
?>  

这是我的XCODE代码

this is my XCODE code

@IBAction func iniciarSesion(_ sender: UIButton) {

       guard
           let emailText = Email.text, !emailText.isEmpty,
           let contrasenaText = Contrasena.text, !contrasenaText.isEmpty else
       {
           displayAlert(title: "Información Faltante", message: "Debes porporcionar un correo y contraseña")
           return
       }

       let myURL = URL(string: "hehe")
       var request = URLRequest(url: myURL!) rather than `NSMutableURLRequest`
       request.httpMethod = "POST"
       let posString = "Email=\(emailText)&Password=\(contrasenaText)"
       request.httpBody = posString.data(using: .utf8)
       let task = URLSession.shared.dataTask(with: request) {
           data, response, error in

           if let error = error {
               print("error=\(error)")
               return
           }

           guard let data = data else {
               print("Something wrong")
               return
           }

           print(String(data: data, encoding: .utf8) as Any)
           do {

               let json = try JSONSerialization.jsonObject(with: data)


               if let parseJSON = json  as? [String: Any]{
                   guard let resultValue = parseJSON["status"] as? String else {
                       print("No `status` in result")
                       return
                   }
                   print("message: \(resultValue) ")

                   if (resultValue == "success") {

                       UserDefaults.standard.set(true, forKey: "isUserLoggedIn")
                       UserDefaults.standard.synchronize()
                   } else {
                      cases
                   }
               } else {

                   print("Result JSON is ill-formatted:", json)
               }
           } catch let err {
               print(err)

           }
       }
       task.resume()
   }

我尝试了什么：



解析JSON是有效的但我猜我的错误是PHP我不知道该怎么办

What I have tried:

parsing JSON is valid but my i guess my error is PHP i don't know what to do

推荐答案

con = mysqli_co nnect（ cantshowthis， 和合， 对不起， ：C）;

//检查连接
if（mysqli_connect_errno（））
{
echo无法连接到MySQL：。 mysqli_connect_error（）;
}

//这个SQL语句从'Registro'表中选择ALL

con=mysqli_connect("cantshowthis","hehe","sorry",":c"); // Check connection if (mysqli_connect_errno()) { echo "Failed to connect to MySQL: " . mysqli_connect_error(); } // This SQL statement selects ALL from the table 'Registro'

sql =SELECT * FROM Registro;

//检查是否有结果
if（

sql = "SELECT * FROM Registro"; // Check if there are results if (

result = mysqli_query（

result = mysqli_query(

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