JSON数据没有从Android应用程序中的服务器获取

所以我刚开始玩JSON和Volley。我已经使用XAMMP创建了一个服务器，并添加了生成JSON数据所需的php脚本。它在浏览器中显示正常。但是，在模拟器上运行应用程序时。数据永远不会被检索，并且进度对话框会无限载入，所以应用程序永远不会崩溃。我采用的代码如下。我错过了什么？JSON数据没有从Android应用程序中的服务器获取

public class Config { 


public static final String DATA_URL2= "http://10.0.2.2:8080/webservice/index.php"; 
public static final String KEY_ID = "ID"; 
public static final String KEY_NAME = "Name"; 
public static final String KEY_SURNAME = "Surname"; 
public static final String JSON_ARRAY = "result";} 

public class MainActivity extends AppCompatActivity implements View.OnClickListener { 

private EditText editTextId; 
private Button buttonGet; 
private TextView textViewResult; 

private ProgressDialog loading; 

@Override 
protected void onCreate(Bundle savedInstanceState) { 
    super.onCreate(savedInstanceState); 
    setContentView(R.layout.activity_main); 

    editTextId = (EditText) findViewById(R.id.editTextId); 
    buttonGet = (Button) findViewById(R.id.buttonGet); 
    textViewResult = (TextView) findViewById(R.id.textViewResult); 

    buttonGet.setOnClickListener(this); 
} 

private void getData() { 

    loading = ProgressDialog.show(this,"Please wait...","Fetching...",false,false); 

    String url = Config.DATA_URL2.toString(); 

    StringRequest stringRequest = new StringRequest(url, new Response.Listener<String>() { 
     @Override 
     public void onResponse(String response) { 
      loading.dismiss(); 
      showJSON(response); 
     } 
    }, 
      new Response.ErrorListener() { 
       @Override 
       public void onErrorResponse(VolleyError error) { 

       } 
      }); 

    RequestQueue requestQueue = Volley.newRequestQueue(this); 
    requestQueue.add(stringRequest); 
} 

private void showJSON(String response){ 
    String id=""; 
    String name=""; 
    String surname = ""; 
    try { 
     JSONObject jsonObject = new JSONObject(response); 
     JSONArray result = jsonObject.getJSONArray(Config.JSON_ARRAY); 
     JSONObject collegeData = result.getJSONObject(0); 
     id = collegeData.getString(Config.KEY_ID); 
     name = collegeData.getString(Config.KEY_NAME); 
     surname = collegeData.getString(Config.KEY_SURNAME); 
    } catch (JSONException e) { 
     e.printStackTrace(); 
    } 
    textViewResult.setText("ID:\t"+id+"\nName:\t" +name+ "\nSurname:\t"+ surname); 
} 

@Override 
public void onClick(View v) { 
    getData(); 
}}

这里是我的堆栈跟踪：

02-14 00:15:19.215 22098-22098/? I/art: Not late-enabling -Xcheck:jni (already on) 
02-14 00:15:19.807 22098-22125/za.co.volleydemo D/OpenGLRenderer: Render dirty regions requested: true 
02-14 00:15:19.839 22098-22098/za.co.volleydemo D/Atlas: Validating map... 
02-14 00:15:19.902 22098-22106/za.co.volleydemo I/art: Background partial concurrent mark sweep GC freed 1945(127KB) AllocSpace objects, 0(0B) LOS objects, 52% free, 921KB/1945KB, paused 5.021ms total 25.227ms 
02-14 00:15:19.958 22098-22125/za.co.volleydemo I/OpenGLRenderer: Initialized EGL, version 1.4 
02-14 00:15:20.046 22098-22125/za.co.volleydemo D/OpenGLRenderer: Enabling debug mode 0 
02-14 00:15:20.073 22098-22125/za.co.volleydemo W/EGL_emulation: eglSurfaceAttrib not implemented 
02-14 00:15:20.073 22098-22125/za.co.volleydemo W/OpenGLRenderer: Failed to set EGL_SWAP_BEHAVIOR on surface 0xae2c5a40, error=EGL_SUCCESS 
02-14 00:15:23.058 22098-22182/za.co.volleydemo E/Volley: [275] BasicNetwork.performRequest: Unexpected response code 401 for http://10.0.2.2:8080 
02-14 00:15:23.076 22098-22125/za.co.volleydemo W/EGL_emulation: eglSurfaceAttrib not implemented 
02-14 00:15:23.077 22098-22125/za.co.volleydemo W/OpenGLRenderer: Failed to set EGL_SWAP_BEHAVIOR on surface 0xa609f700, error=EGL_SUCCESS 
02-14 00:15:23.220 22098-22182/za.co.volleydemo E/Volley: [275] BasicNetwork.performRequest: Unexpected response code 401 for http://10.0.2.2:8080

此外...这是我的PHP脚本

<?php 

define('HOST','localhost'); 
    define('USER','ruchen'); 
    define('PASS','lollatjie'); 
    define('DB','webservice1'); 

$con = mysqli_connect(HOST,USER,PASS,DB); 

$sql = "select * from users where surname = 'Miller'"; 

$res = mysqli_query($con,$sql); 

$result = array(); 

while($row = mysqli_fetch_array($res)){ 
array_push($result, 
array('id'=>$row[0], 
'name'=>$row[1], 
'surname'=>$row[2] 
)); 
} 

echo json_encode(array("result"=>$result)); 

mysqli_close($con); 
?>

最后，通过我的互联网浏览器由脚本生成的JSON数据...

{"result":[{"id":"7","name":"Bob","surname":"Miller"}]}

来源

2016-02-13 R.Wyngaard

401意味着你没有被授权这个网址。

从这个链接 http codes

10.4.2 401未授权

请求需要用户验证。响应必须包含一个WWW-Authenticate头域（14.47节），其中包含一个适用于所请求资源的挑战。客户端可以用适当的授权标题字段重复请求（14.8节）。如果请求已经包含授权凭证，那么401响应表明授权已被拒绝这些凭据

来源

2016-02-13 22:45:35 omer727

我不知道为什么你会得到错误代码401，但我注意到你的代码中有一些奇怪的东西。你为什么要做一个字符串请求而不是JsonObject请求？

毕竟你从php服务获取JSON数据？

将它更改为JsonObjectRequest，因为Json的开头是一个名为“result”的Object，它包含一个数组。

我做同样的事情在我的应用程序和我的排球下载看起来像这样：

theEvents = new ArrayList<Event>(); 
    requestQueue = Volley.newRequestQueue(getActivity()); 
    JsonObjectRequest jsonObjectRequest = new JsonObjectRequest 
      (Request.Method.GET, urlService, null, new Response.Listener<JSONObject>() { 

       @Override 
       public void onResponse(JSONObject response) { 

        try { 


         id = response.getString(Config.KEY_ID); // response is now the actual object called "result" from your JSON data. 
          // do something with the ID 
         } 
.........etc

来源

2016-02-13 22:57:45 iBobb

JSON数据没有从Android应用程序中的服务器获取

回答

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