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android应用程序无法运行并显示以下错误_Android - Fatal编程技术网

android应用程序无法运行并显示以下错误

android应用程序无法运行并显示以下错误,android,Android,我在连接到本地主机时出错,请帮助我的代码是 public class MainActivity extends Activity implements OnClickListener{ /** Called when the activity is first created. */ TextView resultView; HttpClient client; JSONObject json; String Dat; Button btn; InputStream is=n

我在连接到本地主机时出错,请帮助我的代码是

  public class MainActivity extends Activity implements OnClickListener{
/** Called when the activity is first created. */
  TextView resultView;
   HttpClient client;
   JSONObject json;
String Dat;
Button btn;
InputStream is=null;
EditText name,roll_no;
String result="";
StringBuilder sb=null;
JSONArray jArray;
@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.main);
    resultView = (TextView) findViewById(R.id.tv1);
    submit();
    btn.setOnClickListener(this);

}
public void onClick(View v){
parse();

}
private String submit() {
     ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
        //http post
        try{
             HttpClient httpclient = new DefaultHttpClient();
             HttpPost httppost = new HttpPost("http://10.0.2.2:8081/onlineres/login");
             httppost.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());
            }
        try{
              BufferedReader reader = new BufferedReader(new InputStreamReader(is,"iso-8859-1"),8);
               sb = new StringBuilder();
               sb.append(reader.readLine() + "\n");

               String line="0";
               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());
                }
    return result;
  }
    public void parse(){
    String roll,name;
    try{
          jArray  = new JSONArray(result);
          JSONObject json_data=null;
          for(int i=0;i<jArray.length();i++){
                 json_data = jArray.getJSONObject(i);
                 name=json_data.getString("Name");
                 roll =json_data.getString("Roll_no");
             }
          }
          catch(JSONException e1){
              Toast.makeText(getBaseContext(), "No City Found"     ,Toast.LENGTH_LONG).show();
          } catch (ParseException e1) {
                e1.printStackTrace();
        }
}
}

请帮我解决这个问题,我猜问题是你调用了
btn.setOnClickListener(这个)


在您的活动中,我没有看到“btn”已初始化。

您的
btn
变量尚未初始化。在添加onclick侦听器之前,必须从视图中获取它

btn = (Button) findViewById(R.id.btnID);
这需要是你的按钮的id,所以只要用你的按钮id替换它就行了


编辑:忘记播放:)对不起。

您没有初始化按钮。将btn初始化为:

 ........
    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.main);
        btn= (Button) findViewById(R.id.Your_button_id);
        ......

如果按钮位于布局XML中,请查看其他响应。它应该与您在onCreate()中初始化的resultView文本视图非常相似。非常感谢,伙计,它工作得很好。小错误很糟糕,谢谢,我能帮上忙。如果这有效并解决了问题,请接受答案,以便将来帮助他人。
 ........
    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.main);
        btn= (Button) findViewById(R.id.Your_button_id);
        ......