Android-无法从ListView调用setOnItemClickListener

Question

我一直在关注Udemy上的一些Android开发教程，这些教程非常有帮助，但是现在我遇到了导师代码的问题。 不知道这是由于更新的SDK还是其他原因引起的，但是该代码在他的视频中并未出现错误。

package com.example.app;

import android.app.ListActivity;
import android.support.v7.app.ActionBarActivity;
import android.support.v7.app.ActionBar;
import android.support.v4.app.Fragment;
import android.os.Bundle;
import android.view.LayoutInflater;
import android.view.Menu;
import android.view.MenuItem;
import android.view.View;
import android.view.ViewGroup;
import android.os.Build;
import android.widget.AdapterView;
import android.widget.ArrayAdapter;
import android.widget.ListView;
import android.widget.TextView;
import android.widget.Toast;
import android.widget.AdapterView.OnItemClickListener;

public class MainActivity extends ListActivity {

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        String[]countries=getResources().getStringArray(R.array.countries);
        this.setListAdapter(new ArrayAdapter<String>(
                this,
                R.layout.list_element,
                R.id.countryName,
                countries));
    }

    ListView listview = getListView();

    listview.setOnItemClickListener(new AdapterView.OnItemClickListener() {
        public void onItemClick(AdapterView<?>  parent, View clickView, int position, long id) {
            String country = countries[position];
            Toast.makeText(MainActivity.this, String.format("%s was chosen.", country),
                    Toast.LENGTH_SHORT).show();
        }
    });


    @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;
    }

    @Override
    public boolean onOptionsItemSelected(MenuItem item) {
        // Handle action bar item clicks here. The action bar will
        // automatically handle clicks on the Home/Up button, so long
        // as you specify a parent activity in AndroidManifest.xml.
        int id = item.getItemId();
        if (id == R.id.action_settings) {
            return true;
        }
        return super.onOptionsItemSelected(item);
    }

    /**
     * A placeholder fragment containing a simple view.
     */
    public static class PlaceholderFragment extends Fragment {

        public PlaceholderFragment() {
        }

        @Override
        public View onCreateView(LayoutInflater inflater, ViewGroup container,
                Bundle savedInstanceState) {
            View rootView = inflater.inflate(R.layout.fragment_main, container, false);
            return rootView;
        }
    }

}

错误从此处开始，出现“无法解析符号”错误。

listview.setOnItemClickListener

并且在整个包含循环中还会继续出现其他错误。

Answer 1

在onCreate（）中编写该代码...您正在编写会产生语法错误的方法之外...

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    String[] countries = getResources().getStringArray(R.array.countries);
    this.setListAdapter(new ArrayAdapter<String>(this,
            R.layout.list_element, R.id.countryName, countries));

    ListView listview = getListView();

    listview.setOnItemClickListener(new AdapterView.OnItemClickListener() {
        public void onItemClick(AdapterView<?> parent, View clickView,
                int position, long id) {
            String country = countries[position];
            Toast.makeText(MainActivity.this,
                    String.format("%s was chosen.", country),
                    Toast.LENGTH_SHORT).show();
        }
    });
}

Answer 2

这行代码

listview.setOnItemClickListener

不能直接放置在类中，必须将其放置在某些函数的范围内，例如onCreate()

Answer 3

不要使用这个：

 listview.setOnItemClickListener(new AdapterView.OnItemClickListener() 
{
    public void onItemClick(AdapterView<?>  parent, View clickView, int position, long id) 
  {
   .
   .   //your code
   .
  }
}

代替@Override方法onListItemClick方法扩展的类ListActivity的OnCreate外

  @Override
  protected void onListItemClick(ListView l, View v, int position, long id) 
  {
   .
   .  //your code
   .    
  }

Answer 4

您需要在onCreate（）方法中编写该代码。