android連接mysql

Question

即時通訊嘗試在Android應用程序中連接mysql。以下是我的代碼。一旦我運行代碼，我得到'解析數據時出錯org.json.JSONException：在'字符0處輸入結束'錯誤。我用this教程

代碼

public class Test extends Activity{ 
    /** Called when the activity is first created. */ 
    @Override 
     public void onCreate(Bundle savedInstanceState) { 
      super.onCreate(savedInstanceState); 
      //setContentView(R.layout.mainabout); 
      String result = ""; 
      //the year data to send 
      ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(); 
      nameValuePairs.add(new BasicNameValuePair("year","1980")); 
      InputStream is = null;    
      //http post 
      try{ 
        HttpClient httpclient = new DefaultHttpClient(); 

        HttpGet httppost = new HttpGet("http://www.pherma.net84.net/admin/getAllPeopleBornAfter.php"); 
       // httppost.s//setEntity(new UrlEncodedFormEntity(nameValuePairs)); 
        HttpResponse response = httpclient.execute(httppost); 
        HttpEntity entity = response.getEntity(); 
        is = entity.getContent(); 
      }catch(Exception e){ 
        Log.e("log_tag", "Error in http connection "+e.toString()); 
      } 
      //convert response to string 
      try{ 
        BufferedReader reader = new BufferedReader(new InputStreamReader(is,"iso-8859-1"),8); 
        StringBuilder sb = new StringBuilder(); 
        String line = null; 
        while ((line = reader.readLine()) != null) { 
          sb.append(line + "\n"); 
        } 
        is.close(); 

        result=sb.toString(); 
      } 
      catch(Exception e){ 
        Log.e("log_tag", "Error converting result "+e.toString()); 
      } 

      try 
      { 
       JSONArray jArray = new JSONArray(result); 
       for(int i=0;i<jArray.length();i++){ 
         JSONObject json_data = jArray.getJSONObject(i); 
         Log.i("log_tag","id: "+json_data.getInt("id")+ 
           ", name: "+json_data.getString("name")+ 
           ", sex: "+json_data.getInt("sex")+ 
           ", birthyear: "+json_data.getInt("birthyear") 
        ); 
       } 

      } 
      catch(JSONException e){ 
       Log.e("log_tag", "Error parsing data "+e.toString()); 
      } 
     } 
    } 

logcat的

php 

<?php 
mysql_connect(".com","a4055820_root",""); 
mysql_select_db("a4055820_pherma"); 

$q=mysql_query("SELECT * FROM people WHERE birthyear>'".$_REQUEST['year']."'"); 
while($e=mysql_fetch_assoc($q)) 
     $output[]=$e; 

print(json_encode($output)); 

mysql_close(); 
?> 

Answer 1

問題是我忘記了在清單中提供Internet訪問權限。

Answer 2

代碼似乎在第一個try catch塊意味着其餘的工作不會失敗。

您確定您創建的php文件的正確URL？

你的php文件是什麼樣的？

Answer 3

可信的和/或官方渠道

經歷以下鏈接：

1. Error in http connection
2. org.json.JSON Exception : End of input at character 0
3. http://androidforums.com/application-development/217771-parsing-data-using-json-php-but-receiving-error-parsing-data-org-json-exception.html