將JSONobjects從Android發佈到PHP腳本

我期待從android應用程序獲取一些數據，將其發佈到PHP腳本，然後將其寫入到PostGreSQL數據庫中。我有一些困難，有誰能解釋爲什麼數據沒有被傳輸。我不斷收到很多StrictMode違規。我希望當用戶點擊應用上的「上傳」時，整個過程就會自動完成，數據會自動寫入我的PGSQL服務器。將JSONobjects從Android發佈到PHP腳本

Android應用

protected void syncToWebService(final JSONObject json){   
     HttpClient httpclient = new DefaultHttpClient(); 
     HttpConnectionParams.setConnectionTimeout(httpclient.getParams(), 10000); 
     HttpResponse response; 
     String httppost = "http://users.aber.ac.uk/dwd/mfb/php/jsonscript.php";  

     try 
     { 
      HttpPost post = new HttpPost(httppost); 

      Bitmap bitmapOrg = BitmapFactory.decodeResource(getResources(), i); 
      ByteArrayOutputStream bao = new ByteArrayOutputStream(); 
      bitmapOrg.compress(Bitmap.CompressFormat.JPEG, 90, bao); 
      byte[] ba = bao.toByteArray(); 
      String ba1=Base64.encodeToString(ba, i); 

      ArrayList<NameValuePair> postParameters = new ArrayList<NameValuePair>(); 

      postParameters.add(new BasicNameValuePair("photo", ba1.toString())); 
      postParameters.add(new BasicNameValuePair("name", resultName.getText().toString())); 
      postParameters.add(new BasicNameValuePair("description", resultDescription.getText().toString())); 
      postParameters.add(new BasicNameValuePair("latitude", resultLat.getText().toString())); 
      postParameters.add(new BasicNameValuePair("longitude", resultLong.getText().toString())); 
      postParameters.add(new BasicNameValuePair("project", resultProject.getText().toString())); 
      postParameters.add(new BasicNameValuePair("owner", username)); 
      //response = CustomHttpClient.executeHttpPost(httppost, postParameters); 

      post.setEntity(new UrlEncodedFormEntity(postParameters)); 
      response = httpclient.execute(post); 

      /*Checking response variable*/ 
      if(response!=null){ 
       InputStream in = response.getEntity().getContent(); 
      } 
     } 

     catch (Exception e){ 
      e.printStackTrace(); 
     } 
    }

PHP文件

$conn = pg_connect("dbconnection_string"); 

    $jsonString = file_get_contents('php://input'); 
    $jsonObj = json_decode($jsonString, true); 

    if(!empty($jsonObj)) { 

      try { 
        $name = jsonObj['name']; 
        $desc = jsonObj['description']; 
        $latitude = jsonObj['latitude']; 
        $longitude = jsonObj['longitude']; 
        $project = jsonObj['project']; 
        $owner = jsonObj['owner']; 
      } 

    } 
    //decode photo string 
    $photo = $_REQUEST["photo"]; 

    echo $photo; 
    $binary=base64_decode($photo); 
    header('Content-Type: bitmap; charset=utf-8'); 

    $id = pg_query("SELECT * FROM users WHERE email = $owner"); 
    $id_assoc = pg_fetch_assoc($id); 
    $id_res = $id_assoc['u_id']; 

    $res = pg_query("INSERT INTO records (photo, name, description, latitude, longitude, project, owner) VALUES ('$photo', '$name', '$desc', '$latitude', '$longitude', '$project', '$id_res'"); 

    pg_close($conn);

任何幫助您能給將不勝感激。

來源

2013-03-11 WebDevDanno

你在做的UI線程的網絡連接？如是;那麼這就是爲什麼你會得到'StrictMode'違規... – 2013-03-11 12:43:54

在客戶端沒有json ... – njzk2 2013-03-11 12:49:56

我在'onCreate（）'方法中構建'StrictMode'，但網絡連接是在上面的代碼不是嗎？ – WebDevDanno 2013-03-11 12:50:23

改變這一點，你在jsonObj

try { 
       $name = jsonObj['name']; 
       $desc = jsonObj['description']; 
       $latitude = jsonObj['latitude']; 
       $longitude = jsonObj['longitude']; 
       $project = jsonObj['project']; 
       $owner = jsonObj['owner']; 
    }

錯過$到

try { 
       $name = $jsonObj['name']; 
       $desc = $jsonObj['description']; 
       $latitude = $jsonObj['latitude']; 
       $longitude = $jsonObj['longitude']; 
       $project = $jsonObj['project']; 
       $owner = $jsonObj['owner']; 
    }

來源

2013-03-11 12:44:35

什麼是嘗試{}塊，並忽略發生的任何異常？ – 2013-03-11 12:56:29

將JSONobjects從Android發佈到PHP腳本

回答

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