無法使用Java獲取列表項目正確的代碼

我想創建一個Java列表視圖，允許每個列表項目打開一個網址，但我似乎無法獲得正確的代碼。請有人告訴我哪裏出了問題。無法使用Java獲取列表項目正確的代碼

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); 
     } 
    } 

}

來源

2015-10-14 Super Sasquatch

莫非你請詳細說明你得到的錯誤。我想你可能會因爲擴展Activity而得到編譯時錯誤。不是ListActivity並嘗試調用setListAdapter方法。 seContentView（int）也未被調用。 – Gobinath

您有錯誤 - 您的listview在哪裏？修改代碼

1）創建列表視圖 -

ListView myList;

2）在你的onCreate分配此列表：

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

3）設置適配器列表：

myList.setAdapter(adapter);

來源

2015-10-14 17:26:08 Erminesoft

無法使用Java獲取列表項目正確的代碼

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