2015-10-14 71 views
0

我想创建一个Java列表视图,允许每个列表项目打开一个网址,但我似乎无法获得正确的代码。请有人告诉我哪里出了问题。无法使用Java获取列表项目正确的代码

package com.sasquatchapps.hydraquip10.bestofmonsterquest; 

import android.app.Activity; 
import android.content.Intent; 
import android.net.Uri; 

import android.os.Bundle; 

import android.view.View; 
import android.widget.ArrayAdapter; 
import android.widget.ListView; 
import android.widget.Toast; 


public class Season1Activity extends Activity{ 

    private String episodes[] = {"America's Loch Ness Monster","Sasquatch Attack", 
      "Giant Squid Found","Birdzilla","Bigfoot","“Mutant K9","Lions in the Backyard","Gigantic Killer Fish","Swamp Beast","Russia's Killer Apemen","Unidentified Flying Creatures","The Real Hobbit", 
      "Giganto: The Real King Kong","American Werewolf"}; 


    @Override 
    protected void onCreate(Bundle savedInstanceState) { 

     super.onCreate(savedInstanceState); 
     ArrayAdapter<String> adapter = new 
       ArrayAdapter<>(this, 
       android.R.layout.simple_list_item_1, episodes); 
     setListAdapter(adapter); 
    } 

    @Override 
    protected void onListItemClick(ListView l, View v, int position, long id) { 
     Toast.makeText(this, "Item clicked;" + episodes[position], Toast.LENGTH_SHORT).show(); 
     if (position == 0) { 
      Intent intent = new Intent(android.content.Intent.ACTION_VIEW, Uri.parse("https://www.youtube.com/watch?v=o7-RdxrCFAg")); 
      startActivity(intent); 
     } 
    } 

} 
+0

莫非你请详细说明你得到的错误。我想你可能会因为扩展Activity而得到编译时错误。不是ListActivity并尝试调用setListAdapter方法。 seContentView(int)也未被调用。 – Gobinath

回答

0

您有错误 - 您的listview在哪里?修改代码

1)创建列表视图 -

ListView myList; 

2)在你的onCreate分配此列表:

myList = (ListView) findViewById(R.id.my_list); 

3)设置适配器列表:

myList.setAdapter(adapter);