JsonObjectRequest - 无法发送参数

  • Z时代
  • 2024-01-10
  • 分类：问答

我不知道为什么我无法从我的PHP代码中获得正确的响应。对于我的PHP代码，它非常简单。它应该只是返回POST参数。但在我的Android项目中，它为所有参数返回null。我试图在我的iOS项目中测试PHP代码。一切都很完美。它返回["target": zh, "arrayString": <__NSSingleObjectArrayI 0x6000000192a0>(hello world), "source": en]因此，我想也许我的Android项目有一些问题，然后我尝试创建一个新项目并再次执行，但仍然遇到同样的问题。以下是我的新Android项目的步骤。JsonObjectRequest - 无法发送参数

构建摇篮

dependencies { 
    compile 'com.android.volley:volley:1.1.0' 
}

AndroidManifest

<uses-permission android:name="android.permission.INTERNET" />

MainActivity

import android.os.Bundle; 
import android.support.v7.app.AppCompatActivity; 
import android.util.Log; 
import com.android.volley.Request; 
import com.android.volley.Response; 
import com.android.volley.VolleyError; 
import com.android.volley.toolbox.JsonObjectRequest; 
import com.android.volley.toolbox.Volley; 
import org.json.JSONArray; 
import org.json.JSONException; 
import org.json.JSONObject; 
public class MainActivity extends AppCompatActivity { 
    @Override 
    protected void onCreate(Bundle savedInstanceState) { 
     super.onCreate(savedInstanceState); 
     setContentView(R.layout.activity_main); 
     String requestString = "http://ddoommaaiinn.com/testPost.php"; 
     JSONObject parameters = new JSONObject(); 
     try { 
      parameters.put("source", "en"); 
      parameters.put("target", "zh"); 
      JSONArray jsonArray = new JSONArray(); 
      jsonArray.put("hello world"); 
      parameters.put("stringArray", jsonArray); 
     } catch (JSONException e) { 
      e.printStackTrace(); 
     } 
     Log.d("JSONParams", parameters.toString()); 
     JsonObjectRequest jsonRequest = new JsonObjectRequest(Request.Method.POST, requestString, parameters, new Response.Listener<JSONObject>() { 
      @Override 
      public void onResponse(JSONObject response) { 
       //TODO: handle success 
       Log.d("successful", response.toString()); 
      } 
     }, new Response.ErrorListener() { 
      @Override 
      public void onErrorResponse(VolleyError error) { 
       error.printStackTrace(); 
       //TODO: handle failure 
      } 
     }); 
     Volley.newRequestQueue(this).add(jsonRequest); 
    } 
}

调试

12-27 08:47:19.994 4363-4363/com.pakhocheung.testjsonobjectrequest D/JSONParams: {"source":"en","target":"zh","stringArray":["hello world"]} 
12-27 08:47:20.050 4363-4395/com.pakhocheung.testjsonobjectrequest D/NetworkSecurityConfig: No Network Security Config specified, using platform default 
12-27 08:47:20.079 4363-4397/com.pakhocheung.testjsonobjectrequest D/OpenGLRenderer: HWUI GL Pipeline 
12-27 08:47:20.885 4363-4397/com.pakhocheung.testjsonobjectrequest I/zygote: android::hardware::configstore::V1_0::ISurfaceFlingerConfigs::hasWideColorDisplay retrieved: 0 
12-27 08:47:20.886 4363-4397/com.pakhocheung.testjsonobjectrequest I/OpenGLRenderer: Initialized EGL, version 1.4 
12-27 08:47:20.886 4363-4397/com.pakhocheung.testjsonobjectrequest D/OpenGLRenderer: Swap behavior 1 
12-27 08:47:20.886 4363-4397/com.pakhocheung.testjsonobjectrequest W/OpenGLRenderer: Failed to choose config with EGL_SWAP_BEHAVIOR_PRESERVED, retrying without... 
12-27 08:47:20.886 4363-4397/com.pakhocheung.testjsonobjectrequest D/OpenGLRenderer: Swap behavior 0 
12-27 08:47:20.892 4363-4397/com.pakhocheung.testjsonobjectrequest D/EGL_emulation: eglCreateContext: 0xa6f859e0: maj 2 min 0 rcv 2 
12-27 08:47:20.895 4363-4397/com.pakhocheung.testjsonobjectrequest D/EGL_emulation: eglMakeCurrent: 0xa6f859e0: ver 2 0 (tinfo 0xb050d880) 
12-27 08:47:20.998 4363-4397/com.pakhocheung.testjsonobjectrequest D/EGL_emulation: eglMakeCurrent: 0xa6f859e0: ver 2 0 (tinfo 0xb050d880) 
12-27 08:47:21.279 4363-4363/com.pakhocheung.testjsonobjectrequest D/successful: {"source":null,"target":null,"arrayString":null}

PHP代码

<?php 
$source = $_POST['source']; 
$target = $_POST['target']; 
$arrayString = $_POST['stringArray']; 
echo json_encode(array('source'=>$source,'target'=>$target,'arrayString'=>$arrayString)); 
?>

回答：

使用JsonObjectRequest时，所需数据将作为（json）字符串在php://input中找到。为了获得这些数据做：

<?php 
$postArray = json_decode(file_get_contents('php://input'), true); 
// now you access the values nearly as before: 
$source = $postArray['source']; 
$target = $postArray['target']; 
$arrayString = $postArray['arrayString']; 
echo json_encode(array('source'=>$source,'target'=>$target,'arrayString'=>$arrayString)); 
?>

回答：

看起来像你$ _POST不为测试试工作

将硬编码值

$source = 'test'; 
$target = 'test-target'; 
$arrayString = 'test-arrayString'; 
echo json_encode(array('source'=>$source,'target'=>$target,'arrayString'=>$arrayString));

它应该给下面的JSON

{"source":"test","target":"test-target","arrayString":"test-arrayString"}

请加如果有什么缺失

以上是 JsonObjectRequest - 无法发送参数 的全部内容， 来源链接： utcz.com/qa/265727.html

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