Android首次展示“新”形象

时间:2012-04-09 17:01:22

标签: android

我已经部署了我的应用程序并且正在使用中。每个月左右我都会更新它并添加新功能。我想仅在用户使用更新的应用时第一次显示“新”图像。我怎么才能只展示一次?我应该从哪里开始?

2 个答案:

答案 0 :(得分:2)

也许这样的事情可能有助于解决您的问题:

public class mActivity extends Activity {
  @Overrride
  public void onCreate(Bundle savedInstanceState) {
      super.onCreate(savedInstanceState);
      this.setContentView(R.id.layout);

      // Get current version of the app
      PackageInfo packageInfo = this.getPackageManager()
          .getPackageInfo(getPackageName(), 0);
      int version = packageInfo.versionCode;

      SharedPreferences sharedPreferences = this.getPreferences(MODE_PRIVATE);
      boolean shown = sharedPreferences.getBoolean("shown_" + version, false);

      ImageView imageView = (ImageView) this.findViewById(R.id.newFeature);
      if(!shown) {
          imageView.setVisibility(View.VISIBLE);

          // "New feature" has been shown, then store the value in preferences
          SharedPreferences.Editor editor = sharedPreferences.edit();
          editor.put("shown_" + version, true);
          editor.commit();
      } else
          imageView.setVisibility(View.GONE);
  }

下次运行应用程序时,图像将不会显示。

答案 1 :(得分:0)

如果您有可用于存储图像的服务器,请将其放在那里并在更新时下载。

简单的方法是使用sharedprefernece来保存当前的应用程序版本,然后在打开时对其执行检查。如果它返回的版本与存储版本不同,则会运行图像下载程序并以所需方式显示它。 :)

这是我在我的一个应用程序中使用的示例,将ImageView声明放在您的oncreate中,并在那里的某处提供方法,以及您的好处:)

要记住的另一件事是,您可以使用图片托管服务商(如imgur)为您的客户托管“应用更新”图像,并且由于您将定期更新,因此您不必担心图像被删除它仍然在使用(如果你保持应用程序更新)

    private final static String URL = "http://hookupcellular.com/wp-content/images/android-app-update.png";

    ImageView imageView = new ImageView(this);
    imageView.setImageBitmap(downloadImage(URL));


private Bitmap downloadImage(String IMG_URL)
{
    Bitmap bitmap = null;
    InputStream in = null;        
    try {
        in = OpenHttpConnection(IMG_URL);
        bitmap = BitmapFactory.decodeStream(in);
       in.close();
 } catch (IOException e1) {
   e1.printStackTrace();
 }
 return bitmap;                
 }

private static InputStream OpenHttpConnection(String urlString) throws IOException
{
    InputStream in = null;
    int response = -1;

    URL url = new URL(urlString); 
    URLConnection conn = url.openConnection();

    if (!(conn instanceof HttpURLConnection))                     
        throw new IOException("Not an HTTP connection");

    try
    {
        HttpURLConnection httpConn = (HttpURLConnection) conn;
        httpConn.setAllowUserInteraction(false);
        httpConn.setInstanceFollowRedirects(true);
        httpConn.setRequestMethod("GET");
        httpConn.connect(); 

        response = httpConn.getResponseCode();                 
        if (response == HttpURLConnection.HTTP_OK) {
            in = httpConn.getInputStream();                                 
        }                     
    }
    catch (Exception ex)
    {
        throw new IOException("Error connecting to " + URL);
    }
    return in;     
}

希望这会有所帮助:D