獲取「[]」作爲PHP服務器應答，應該返回一個JSON數組

Question

我面臨的問題是，我想從運行PHP和MySQL的服務器返回一個JSON數組到android應用程序。我沒有得到任何錯誤，但是代碼並沒有按照需要給出輸出。執行我的代碼後，我得到以下輸出。

logcat的條目：

09-26 15:53:54.166: I/System.out(32008): Connected 
09-26 15:53:54.226: D/MyActivity(32008): registered 
09-26 15:53:54.266: D/dalvikvm(32008): GC_FOR_ALLOC freed 254K, 2% free 16955K/17240K, paused 10ms, total 10ms 
09-26 15:53:54.266: I/System.out(32008): Process Result started 
09-26 15:53:54.496: I/System.out(32008): 200 
09-26 15:53:54.496: I/System.out(32008): BR REader:java.io.BufferedReader@426d40e0 
09-26 15:53:54.536: I/System.out(32008): String builder:[] 
09-26 15:53:54.536: I/System.out(32008): Result:[] 
09-26 15:53:54.536: I/System.out(32008): Process Result ended 
09-26 15:53:54.566: D/dalvikvm(32008): GC_FOR_ALLOC freed 253K, 2% free 17216K/17500K, paused 13ms, total 14ms 
09-26 15:53:54.586: I/Adreno-EGL(32008): <qeglDrvAPI_eglInitialize:320>: EGL 1.4 QUALCOMM Build: I0404c4692afb8623f95c43aeb6d5e13ed4b30ddbDate: 11/06/13 
09-26 15:53:54.616: D/OpenGLRenderer(32008): Enabling debug mode 0 

和源代碼是：

protected void onCreate(Bundle savedInstanceState) { 
    .................... 
    .................... 
    syncDatabases = new SyncDatabases(); 

    try { 
     syncDatabases.execute().get(); 
    } catch (InterruptedException e) { 
     e.printStackTrace(); 
    } catch (ExecutionException e) { 
     e.printStackTrace(); 
    } 
} 

public class SyncDatabases extends AsyncTask<Void, Void, String> { 

    String result=null; 

    @Override 
    protected String doInBackground(Void... String) { 
     try { 
      System.out.println("Process Result started"); 
      String result1=getJSONUrl(uploadDetailServerUri); 
      System.out.println("Result:"+result1); 

      JSONArray json=new JSONArray(result1); 
      for (int i = 0; i < json.length(); i++) { 

      JSONObject jsonOb=json.getJSONObject(i); 
      System.out.println("JSON Colection:"+jsonOb.toString()); 
      String imageName = jsonOb.getString("imagename"); 
      String status = jsonOb.getString("status"); 
      System.out.println("ImageName:"+imageName+"Status:"+status); 
      } 
      System.out.println("Process Result ended"); 
      } catch (Exception e) { 
      e.printStackTrace(); 
      } 
     return null; 
    } 

    public String getJSONUrl(String url) { 
     StringBuilder str = new StringBuilder(); 
     HttpClient client = new DefaultHttpClient(); 
     // HttpGet httpGet = new HttpGet(url); 

     HttpPost httpPost = new HttpPost(url); 
     // httpPost.setHeader(HTTP.CONTENT_TYPE, "application/json; charset=utf-8"); 

     List<NameValuePair> nameVP = new ArrayList<NameValuePair>(1); 
      nameVP.add(new BasicNameValuePair("mobile_number", phoneNumber)); 
      try { 
      httpPost.setEntity(new UrlEncodedFormEntity(nameVP)); 
      // httpGet.setParams(new BasicNameValuePair("mobile_number", phoneNumber)); 

      HttpResponse response = client.execute(httpPost); 
      StatusLine statusLine = response.getStatusLine(); 
      int statusCode = statusLine.getStatusCode(); 
      System.out.println(statusCode); 
      if (statusCode == 200) { // Download OK 
      HttpEntity entity = response.getEntity(); 
      InputStream content = entity.getContent(); 
      BufferedReader reader = new BufferedReader(new InputStreamReader(content,"UTF-8")); 
      System.out.println("BR REader:"+reader.toString()); 
      String line; 
      while ((line = reader.readLine()) != null) { 
      str.append(line); 
      } 
      } else { 
      Log.e("Log", "Failed to download file.."); 
      } 
      } catch (ClientProtocolException e) { 
      e.printStackTrace(); 
      } catch (IOException e) { 
      e.printStackTrace(); 
      } 
      System.out.println("String builder:"+str.toString()); 
      return str.toString(); 
    } 


    } 

而PHP代碼是：

<?php 
include ('mysqlconnection.php'); 
$mobile = $_POST['mobile_number']; 
$output = array(); 
$query="select name,status from content;"; 
$sql = mysql_query($query); 
while($row = mysql_fetch_assoc($sql)) { 
    $output[]= array('imagename' => $row['name'], 'status' => $row['status']); 
} 
echo json_encode($output); 
?> 

當瀏覽器發送httprequest時，我們在瀏覽器中獲得了所需的輸出（JSON數組）。

我不知道爲什麼即時通訊只獲取「[]」作爲logcat中的輸出而不是打印JSONArray值。幫助我，謝謝你。

Answer 1

我覺得您使用保留字，改變你的查詢

$query="select name,status from content;"; 

到

$query="select `name`,`status` from content";