ListView不起作用

时间:2012-02-13 15:47:27

标签: android eclipse

我正在尝试在一个喜欢按钮的页面中创建列表视图,     但是当我调试应用程序时,它只显示一个带有名称的黑页。     我有2个java     一个xml到主页面     和一个mainfest.xml

第1页:

package tm.andrioid.com;
import android.app.Activity;
import android.app.Dialog;
import android.app.ListActivity;
import android.os.Bundle;
import android.view.Menu;
import android.view.MenuInflater;
import android.view.View;
import android.widget.Button;
import android.app.Activity;
import android.os.Bundle;
import android.content.Intent;
import android.widget.Button;
public class indexActivity extends Activity
{
public void onCreate(Bundle icicle)
{
  super.onCreate(icicle);
  setContentView(R.layout.main);  
  Button bprodukter = (Button) findViewById(R.id.bprodukter);
  bprodukter.setOnClickListener(new View.OnClickListener() {
     public void onClick(View arg0) {
     Intent i = new Intent(indexActivity.this, produkter.class);
     startActivity(i);     
     } 
  });
 }``
}

第2页java:

 tm.andrioid.com;  
import android.app.Activity;
import android.app.Dialog;
import android.app.ListActivity;
import android.content.Intent;
import android.os.Bundle;
import android.view.View;
import android.widget.ArrayAdapter;
import android.widget.ListView;
import android.app.Activity;
import android.os.Bundle;
import android.widget.Button;
public class produkter extends ListActivity {
      String classes[] = { "indexActivity", "Data", "Intim", "Hörlurar", "Prylar",        

    "Mat", "Sex", "Sju"};
        @Override
        protected void onCreate(Bundle savedInstanceState) {
            // TODO Auto-generated method stub
         super.onCreate(savedInstanceState);
         setListAdapter(new ArrayAdapter<String>(produkter.this,          

     android.R.layout.simple_list_item_1));
      }
      @Override
        protected void onListItemClick(ListView l, View v, int position, long id) {
            // TODO Auto-generated method stub
            super.onListItemClick(l, v, position, id);
            String saker = classes[position];
            try{
            Class ourClass = Class.forName("tm.android.com." + saker);
            Intent ourIntent = new Intent(produkter.this, ourClass);    
            startActivity(ourIntent);
            }catch(ClassNotFoundException e){
                e.printStackTrace();
            }
        }       

}

main xml:

<?xml version="1.0" encoding="utf-8"?>
<manifest xmlns:android="http://schemas.android.com/apk/res/android"
package="tm.andrioid.com"
android:versionCode="1"
android:versionName="1.0" >
<uses-sdk android:minSdkVersion="15" />
<application
    android:icon="@drawable/tmm"
    android:label="@string/app_name" >
    <activity
        android:name=".splash"
        android:label="@string/app_name" >
          <intent-filter>
            <action android:name="android.intent.action.MAIN" />
             <category android:name="android.intent.category.LAUNCHER" />
        </intent-filter>            
          </activity>
    <activity
        android:name=".indexActivity"
        android:label="@string/app_name" >
        <intent-filter>
            <action android:name="tm.android.com.INDEXACTIVITY" />
<category android:name="android.intent.category.DEFAULT" />
          </intent-filter>
    </activity>
    <activity
        android:name=".produkter"
        android:label="@string/app_name" >
        <intent-filter>
            <action android:name="tm.android.com.PRODUKTER" />
          <category android:name="android.intent.category.DEFAULT" />
        </intent-filter>
    </activity>
 </application>
</manifest>

1 个答案:

答案 0 :(得分:0)

替换

setListAdapter(new ArrayAdapter<String>(produkter.this,          

     android.R.layout.simple_list_item_1));

用这个:

setListAdapter(new ArrayAdapter<String>(produkter.this,          

     android.R.layout.simple_list_item_1, classes));

让我知道这是否有效。

相关问题