无法在未调用Looper.Prepare()3的线程内创建处理程序

Question

public class AddStudentActivity extends AppCompatActivity {

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_add_student);
}

public void add2(View view){
    //create 3 editText  ,  1 Textview
    EditText et_name, et_age;
    RadioButton rb_male;
    TextView tv_msg;

    //bind with xml widget
    et_name = (EditText)findViewById(R.id.et_name);
    et_age = (EditText)findViewById(R.id.et_age);
    rb_male = (RadioButton)findViewById(R.id.rb_male);
    tv_msg = (TextView)findViewById(R.id.tv_msg);
    //retrieve values
    String name = et_name.getText().toString();
    String age = et_age.getText().toString();
    String gender = "";
    if (rb_male.isChecked())
        gender = "male";
    else
        gender = "female";
    //call php
    new AddStudent(this, tv_msg).execute(name, age, gender);
   }
}

class AddStudent extends AsyncTask<String, Void, String> {
private Context context;
//CHANGE HERE....ADD PARAMATER
TextView tv_msg;

public AddStudent(Context context,TextView tv_msg) {
    this.context = context;
    this.tv_msg = tv_msg;

}

@Override
protected String doInBackground(String[] arg) {
    //CHANGE HERE ...... RETRIEVE STRING ARGUMENTS
   String name = arg[0];
    String age = arg[1];
    String gender = arg[2];

    String link;
    String data;
    BufferedReader bufferedReader;
    String result;

    try {
        //CHANGE HERE .... PARAMETERS
          data = "?name=" + URLEncoder.encode(name, "UTF-8");
        data += "&age="+ URLEncoder.encode(age, "UTF-8");
        data += "&gender="+ URLEncoder.encode(gender, "UTF-8");
        //CHANGE HERE .... PHP PAGE
          link = "http://10.0.2.2/fashionstudio/ADD3.php" + data;

        URL url = new URL(link);
        HttpURLConnection con = (HttpURLConnection) url.openConnection();

        bufferedReader = new BufferedReader(new InputStreamReader(con.getInputStream(), "UTF-8"));
        result = bufferedReader.readLine();

        return result;
    } catch (Exception e) {
        Log.d("msg here", e.getMessage());
        Toast.makeText(context, "Error Exception", Toast.LENGTH_SHORT).show();
        return new String("Exception: " + e.getMessage());

    }
}

@Override
protected void onPostExecute(String result) {
    if (result != null) {
        try {
            JSONObject reader = new JSONObject(result);
            //CHANGE HERE... RETRIEVE VALUES
            String msg = reader.getString("msg");
            tv_msg.setText(msg);
        } catch (Exception e) {
            e.printStackTrace();
            Log.d("msg here", e.getMessage());
            Toast.makeText(context, "Error parsing JSON data.", Toast.LENGTH_SHORT).show();
        }
    } else {
        Toast.makeText(context, "Couldn't get any JSON data.", Toast.LENGTH_SHORT).show();
    }
  }
}

我在android中点击添加学生时出错了

Answer 1

你不能

Toast.makeText(context, "Error Exception", Toast.LENGTH_SHORT).show();

在doInbackground方法中。toast.maketext(...)必须在UI线程中执行。