Question

所以我刚刚开始玩JSON和Volley。我使用XAMMP创建了一个服务器，并添加了生成JSON数据所需的php脚本。它在浏览器中很好用。但是，在模拟器上运行应用程序时。永远不会检索数据，并且无休止地加载进度对话框，因此应用程序永远不会崩溃。我采用的代码如下。我错过了什么？

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

另外......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，脚本通过我的互联网浏览器生成的JSON数据......

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

Answer 1

401表示您无权使用此网址。

从这个链接 http codes

10.4.2 401未经授权

请求需要用户身份验证。响应必须包含WWW-Authenticate头字段（第14.47节），其中包含适用于所请求资源的质询。客户端可以使用合适的Authorization头字段重复请求（第14.8节）。如果请求已包含授权凭据，则401响应表示已拒绝授权这些凭据

Answer 2

我不确定为什么会出现错误代码401，但我发现代码中有些奇怪。为什么要发出String请求而不是JsonObject请求？！

毕竟你从php服务中获取JSON数据？

将其更改为JsonObjectRequest，因为Json的开头是一个名为“result”的对象，它包含一个数组。

我在我的应用程序中做同样的事情，我的Volley下载看起来像那样：

O(n^2 + n log_4(n) - n - log_4(n) + 1) = O(n^2).

在Android应用程序中没有从服务器获取JSON数据

2 个答案: