JSONException：java.lang.String類型的值無法轉換爲JSONObject

Question

我盡力搜索解決方案，但仍無法解決問題。請幫助。在這裏我的Java代碼： -

public class MainActivity extends Activity { 

String project_id; 
String id; 
InputStream is=null; 
String result=null; 
String line=null; 
int code = 0; 

public void onCreate(Bundle savedInstanceState) { 
    super.onCreate(savedInstanceState); 
    setContentView(R.layout.activity_main); 

    final EditText e_id  =(EditText) findViewById(R.id.editText1); 
    final EditText e_prjId =(EditText) findViewById(R.id.editText2); 
    Button insert   =(Button) findViewById(R.id.button1); 

    id   = e_id.getText().toString(); 
    project_id = e_prjId.getText().toString(); 
    insert.setOnClickListener(new View.OnClickListener() { 
     public void onClick(View v) { 
      insert(); 

     } 
    }); 
    } 

    public void insert() { 
    final ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(); 

    nameValuePairs.add(new BasicNameValuePair("id",id)); 
    nameValuePairs.add(new BasicNameValuePair("Project_Id",project_id)); 


    new Thread(new Runnable() { 
     public void run() { 
      try { 

       HttpClient httpclient = new DefaultHttpClient(); 
       HttpPost httppost = new HttpPost("http://192.168.0.111/insert.php"); 

       httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs)); 
       HttpResponse response = httpclient.execute(httppost); 
       HttpEntity entity = response.getEntity(); 
       is = entity.getContent(); 
       Log.e("pass 1", "connection success "); 
      } 
      catch(Exception e){ 
       Log.e("Fail 1", e.toString()); 
       Toast.makeText(getApplicationContext(), "Invalid IP Address", 
       Toast.LENGTH_LONG).show(); 
      } 

      try { 
       BufferedReader reader = new BufferedReader(new InputStreamReader(is,"iso-8859-1"),8); 
       StringBuilder sb = new StringBuilder(); 

       while ((line = reader.readLine()) != null){ 
        sb.append(line + "\n"); 
       } 

       is.close(); 
       result = sb.toString(); 
       Log.e("pass 2", "connection success "); 
      } 
      catch(Exception e){ 
       Log.e("Fail 2", e.toString()); 
      }  

      try { 
       Log.i("tagconvertstr", "["+result+"]"); 
       JSONObject json_data = new JSONObject(result); 
       code=(json_data.getInt("code")); 
      } catch (JSONException e) { 
       // TODO Auto-generated catch block 
       e.printStackTrace(); 
      }  

      if(code==1) 
      { 
       Toast.makeText(getBaseContext(), "Inserted Successfully",Toast.LENGTH_SHORT).show(); 
      } 
      else 
      { 
       Toast.makeText(getBaseContext(), "Sorry, Try Again",Toast.LENGTH_LONG).show(); 
      } 
     } 
    }).start(); 



} 

PHP： -

<?php 
$uname='root'; 
$pwd=''; 


$con = new PDO("mysql:host=192.168.0.111;dbname=wktask", $uname, $pwd); 


$ID=$_REQUEST['ID']; 
$Project_Id=$_REQUEST['Project_Id']; 

$flag['code']=0; 

if($r= $con->query("insert into task(ID,Project_Id) values('$ID','$Project_Id')")) 
{ 
    $flag['code']=1; 
} 

echo(json_encode($flag)); 

?> 

我真的不知道什麼是我不斷收到來自JSON異常錯誤錯誤信息的原因。真的很感謝能幫助我。

感謝

Answer 1

小心，PHP關聯數組是區分大小寫的

您發送id：

nameValuePairs.add(new BasicNameValuePair("id",id)); 

這不等於ID

除了這個錯誤，你不檢查你的PHP腳本中的數據，我爲你重寫了它：

$data = array(); 

if(isset($_POST['id'], $_POST['Project_Id']){ 
    $id=$_POST['id']; 
    $project_id=$_POST['Project_Id']; 

    $uname='root'; 
    $pwd=''; 
    $con = new PDO("mysql:host=192.168.0.111;dbname=wktask", $uname, $pwd); 
    $con->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); 
    $stmt = $con->prepare('INSERT INTO task (`ID`, `Project_Id`) values(:id, :project_id)')) 
    $success = $stmt->execute(array(':id'=>$id, ':project_id'=>$project_id)); 

    if($success){ 
     $data['code'] = 1; 
     $data['msg'] = 'INSERT successful'; 
    }else{ 
     $data['code'] = 0; 
     $data['msg'] = 'INSERT Failed'; 
    } 


}else{ 
    $data['code'] = 0; 
    $data['msg'] = 'values are not set'; 
} 

echo(json_encode($data));