Question

这里的奇怪部分是有效的...然后我尝试了一些不起作用的东西，所以回到我原来的代码，这个代码目前无效。在解析数据期间发生错误。

这是我的代码。

    //@SuppressLint("NewApi")
public class MainActivity extends Activity {
    InputStream isr = null;
    String result = "";
    TextView resultView;

    @TargetApi(Build.VERSION_CODES.GINGERBREAD)
    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);
        try {
            result = new httprRequest().execute().get();
        } catch (InterruptedException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        } catch (ExecutionException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
        getData();
        resultView = (TextView) findViewById(R.id.result);


    }

/*
    @Override
    public boolean onCreateOptionsMenu(Menu menu) {
        // Inflate the menu; this adds items to the action bar if it is present.
        getMenuInflater().inflate(R.menu.main, menu);
        return true;
    }
*/
    public void getData(){
        String s = "";

    //parse json data
        try {

               JSONArray jArray = new JSONArray(result);

               for(int i=0; i<jArray.length();i++){
                   JSONObject json = jArray.getJSONObject(i);
                   s = s + 
                           "Name : "+json.getString("first_name")+"\n"+

                           "Mobile Using : "+json.getString("mobile_phone")+"\n\n";
               }

               resultView.setText(s);

           } catch (Exception e) {
            // TODO: handle exception
               Log.e("log_tag", "Error Parsing Data "+e.toString());
           }


    }

private class httprRequest extends AsyncTask<String,Integer,String>{

        @Override
        public String doInBackground(String... params){
            ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
            try
            {
                HttpClient httpclient = new DefaultHttpClient();
                HttpPost httppost = new HttpPost("private");
                httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
                HttpResponse response = httpclient.execute(httppost);
                if(response.getStatusLine().getStatusCode()!=200){
                    Log.d("MyApp", "Server encontered an error.");
                }
            HttpEntity entity = response.getEntity();
            isr = entity.getContent();
            }catch(Exception e){
                Log.e("log_entity", "Error in http connection: "+e.toString());
                }

            //conversion happening here..
        try{
            BufferedReader reader = new BufferedReader(new InputStreamReader(isr,"iso-8859-1"),8);
            StringBuilder sb = new StringBuilder();
            String line = null;
            while ((line = reader.readLine()) != null) {
                    sb.append(line + "\n");
            }
            isr.close();

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

        return result;

        }
    }
}

这是我的php文件输出的内容。

[{"first_name":"CHRIS","mobile_phone":"+278285187"},{"first_name":"IAN","mobile_phone":"+64 (27) 582 58"},{"first_name":"ANDRIES","mobile_phone":"+278257922"}]

这是我的PHP代码

connect();

$sql=mysql_query("SELECT first_name, mobile_phone FROM `admin_contacts`");

$output = array();
while($r = mysql_fetch_assoc($sql)){
    $output[]=$r;

}
echo json_encode($output);

?>

我的JSOn有效。它抛出一个java.lang.NullPointerException .....如果有帮助的话。

谢谢安德鲁......简直不敢相信我没有看到 - .- ......哈哈，你救了我的话

Answer 1

这是无效的json，它周围应该有[]标记为数组

编辑：

看起来实际问题在这里

    getData();
    resultView = (TextView) findViewById(R.id.result);

getData函数在填充之前使用resultView。交换这些线，这应该解决它

我在从php文件转换JSon数据时遇到了解析问题

1 个答案: