如何将 PHP 与 JSON 类型标头连接起来

Posted 2023-04-19

【中文标题】如何将 PHP 与 JSON 类型标头连接起来

【英文标题】：How to connect PHP with JSON type header

【发布时间】：2020-10-11 22:46:21

【问题描述】：

这是我的 php 代码

<?php require_once '../includes/DbOperations.php'; $response = array(); 

if($_SERVER['REQUEST_METHOD']=='POST')
    if(isset($_POST['username']) and isset($_POST['password']) and isset($_POST['email']))
    
        //operate the data further 

        $db = new DbOperations(); 

        $result = $db->createUser($_POST['username'],$_POST['password'],$_POST['email']);
        if($result == 1)
            $response['error'] = false; 
            $response['message'] = "User registered successfully";
        elseif($result == 2)
            $response['error'] = true; 
            $response['message'] = "Some error occurred please try again";          
        elseif($result == 0)
            $response['error'] = true; 
            $response['message'] = "It seems you are already registered, please choose a different email and username";                     
        

    else
        $response['error'] = true; 
        $response['message'] = "Required fields are missing";
    
else
    $response['error'] = true; 
    $response['message'] = "Invalid Request";


echo json_encode($response);
?>

当我使用 JSON 类型标头时它不起作用

String jsonResult = null;
    JSONObject jsonObject = new JSONObject();
    try 
        jsonObject.put("username", "33");
        jsonObject.put("password", "33");
        jsonObject.put("email", "[e-mail]");

        jsonResult = jsonObject.toString();
     catch (JSONException e) 
        // TODO Auto-generated catch block
        e.printStackTrace();
    


    String HitURL = "http://192.xxx.xxx.xxx/android/v1/registerUser.php";
    MediaType MEDIA_PlainMT = MediaType.parse("text/plain; charset=utf-8");
    Request request = new Request.Builder().url(HitURL).post(RequestBody.create(MEDIA_PlainMT, jsonResult)).build();

    OkHttpClient client = new OkHttpClient();
    try 
        TLSSocketFactory tlsSocketFactory=new TLSSocketFactory();
        if (tlsSocketFactory.getTrustManager()!=null) 
            client = new OkHttpClient.Builder()
                    .sslSocketFactory(tlsSocketFactory, tlsSocketFactory.getTrustManager())
                    .build();
        
     catch (KeyManagementException e) 
        e.printStackTrace();
     catch (NoSuchAlgorithmException e) 
        e.printStackTrace();
     catch (KeyStoreException e) 
        e.printStackTrace();
    
    client.newCall(request).enqueue(new Callback() 
        @Override
        public void onFailure(okhttp3.Call call, IOException e) 
            e.printStackTrace();
        

        @Override
        public void onResponse(Call call, final Response response) throws IOException 
            if (!response.isSuccessful()) 
                throw new IOException("Unexpected code " + response);
             else 
                String HitResponse = response.body().string();
                Log.v("/CheckLog/", HitResponse);
            
        
    );

结果

"error":true,"message":"Required fields are missing"

【问题讨论】：

请写下你的发送代码sn-ps

@ASHKARAN 我更新了我的帖子，我使用的代码被剪掉了

【参考方案1】：

您的服务器代码需要提交常规的 application/x-www-form-urlencoded 值（也称为常规 POST 表单提交），但您的客户端正在将您的数据编码为 application/json（JSON 对象）。应该是哪一个？客户端和服务器必须相互一致。

如果您想使用 JSON 作为数据编码标准，您的代码应如下所示：

<?php

require_once '../includes/DbOperations.php';
$response = array(); 

if($_SERVER['REQUEST_METHOD']=='POST')
    $raw_input = file_get_contents('php://input');
    try 
      $input = json_decode($raw_input, TRUE, 512, JSON_THROW_ON_ERROR);
      if (isset($input['username']) and isset($input['password']) and isset($input['email']))
      
          //operate the data further 

          $db = new DbOperations(); 

          $result = $db->createUser($input['username'],$input['password'],$input['email']);
          if($result == 1)
              $response['error'] = false; 
              $response['message'] = "User registered successfully";
          elseif($result == 2)
              $response['error'] = true; 
              $response['message'] = "Some error occurred please try again";          
          elseif($result == 0)
              $response['error'] = true; 
              $response['message'] = "It seems you are already registered, please choose a different email and username";                     
          

      else
          $response['error'] = true; 
          $response['message'] = "Required fields are missing";
      
   catch (Exception $e) 
      $response['error'] = true; 
      $response['message'] = "Invalid Request: " . $e->getMessage();
  
else
    $response['error'] = true; 
    $response['message'] = "Invalid Request";


echo json_encode($response);

【讨论】：

我从你的回答中学到了一些东西，非常感谢你，但是当我使用你的代码时，它在第 10 行说错了，你能帮忙看看缺少什么吗？ , 致命错误：未捕获的错误：不能使用 stdClass 类型的对象作为 C:\AllMyFiles\Featuring\Database\Program\Xampp\htdocs\Android\v1\registerUser.php 中的数组： 10 堆栈跟踪：在 10 行的 C:\AllMyFiles\Featuring\Database\Program\Xampp\htdocs\Android\v1\registerUser.php 中抛出 #0 main

我忘记了 json_decode 默认为你提供对象。所以结果$input 值不能用$input['key'] 检索。我已将 json_decode $assoc 标志更改为 TRUE，现在应该可以正常工作了。或者，您可以改为通过$input->key 获得某个密钥。两者都应该工作。

谢谢它现在正在工作，JSON 对象需要使用 $input->key，如果我将 JSON 对象转换为数组，我需要使用 $input['key']，谢谢您的时间:D