走动作为单独的类

我浪费了很多时间试图写：走动作为单独的类

private void showAbout() { 
    Dialog dialog = new Dialog(Generator.this); 
    dialog.setContentView(R.layout.about); 
    dialog.setTitle(getString(R.string.about)); 
    dialog.setCancelable(true); 

    try { 
     TextView tv_version = (TextView) dialog.findViewById(R.id.tv_version); 
     tv_version.setText("Version number: " + getPackageManager().getPackageInfo(
      getPackageName(), 0).versionName); 
     TextView tv_createdBy = (TextView) dialog 
        .findViewById(R.id.tv_createdBy); 
      tv_createdBy.setText(getString(R.string.made_by)); 
     } catch (NameNotFoundException e) { 
      Log.e(TAG, "showAbout()", e); 
     } finally { 
      dialog.show(); 
     } 
}

在在希望的一类，使我的代码更易读。

我写这样的：

private void showAbout() { 
       About about = new About(); 
       about.show(); 
} 

public class About extends Activity { 
    String TAG = "About"; 
    Dialog dialog; 

    /** 
    * 
    */ 
    public About() { 
     dialog = new Dialog(About.this); 

    } 

    public void show() { 

     dialog.setContentView(R.layout.about); 
     dialog.setTitle(getString(R.string.about)); 
     dialog.setCancelable(true); 

     try { 
      TextView tv_version = (TextView) dialog 
        .findViewById(R.id.tv_version); 
      tv_version 
        .setText("Version number: " 
          + getPackageManager().getPackageInfo(
            getPackageName(), 0).versionName); 
      TextView tv_createdBy = (TextView) dialog 
        .findViewById(R.id.tv_createdBy); 
      tv_createdBy.setText(getString(R.string.made_by)); 
     } catch (NameNotFoundException e) { 
      Log.e(TAG, "showAbout()", e); 
     } finally { 
      dialog.show(); 

     } 
    } 

}

的Bt这是行不通的。它似乎是在创建对话框时崩溃的，但我不知道如何以另一种方式编写它。

任何想法？

来源

2012-01-31 ringkjob

你怎么称呼这项活动？你有没有在androidmanifest.xml中定义活动？ – ariefbayu 2012-01-31 13:52:52

这是一个很好的观点。您如何确定如何定义活动？我并不擅长这一点 - 但是。 – ringkjob 2012-01-31 18:15:26

你想如何显示'about'？在弹出对话框？或在另一个活动？ – ariefbayu 2012-02-01 01:17:08

鉴于您希望以pop-up的方式显示。那么你做错了。将类扩展到Activity将使其成为活动类（用于活动）。您应该这样做：

//class level variable 
private Dialog formDialog = null; 

/** Called when the activity is first created. */ 
@Override 
public void onCreate(Bundle savedInstanceState) { 
    prepareDialog(); 
} 

/** 
* We prepare dialog in it's own method so that we can maintain code separation well. 
* I believe there is another <em>better</em> solution. But, it work for now. 
*/ 
private void prepareDialog() { 
    formDialog = new Dialog(ListDataActivity.this); 
    formDialog.setContentView(R.layout.form); 
    formDialog.setTitle("Form Buku"); 

    // set dialog width to fill_parent 
    LayoutParams formDialogParams = formDialog.getWindow().getAttributes(); 
    formDialogParams.width = LayoutParams.FILL_PARENT; 
    formDialog.getWindow().setAttributes(
      (android.view.WindowManager.LayoutParams) formDialogParams); 

    txtNama = (EditText) formDialog.findViewById(R.id.txtFormNamaPenulis); 
    txtNama.setText("about info"); 


    btnSimpan = (Button) formDialog.findViewById(R.id.btnFormSimpan); 

    btnBack.setOnClickListener(new OnClickListener() { 

     @Override 
     public void onClick(View view) { 
      formDialog.hide(); 
     } 
    }); 
}

请注意，您还需要为此对话框创建新的.xml布局。

最后，您只需拨打formDialog.show();即可显示。这个答案从我的android tutorial中提取并略微修改。

来源

2012-02-01 23:38:19 ariefbayu