我正在使用getBitmap方法显示图像。由于我将其作为方法使用,如果它返回位图,则显示图像,但如果它返回null,则捕获异常。但是,如果输入的网址不正确,它还应该处理FileNotFoundException。如何处理这两个异常并在用户界面上显示?
public Bitmap getBitmap(final String src) {
try {
InputStream stream = null;
URL url = new URL(src);
java.net.URL url = new java.net.URL(src);
URLConnection connection = url.openConnection();
InputStream input = connection.getInputStream();
myBitmaps = BitmapFactory.decodeStream(input);
return myBitmaps;
} catch (IOException e) {
e.printStackTrace();
Log.e("IO","IO"+e);
return null;
}
catch(OutOfMemoryError e1) {
e1.printStackTrace();
Log.e("Memory exceptions","exceptions"+e1);
return null;
}
}
在Activity中,我给了这样的代码:
Bitmap filename=service.getBitmap(url_box.getText().toString());
if(file_name!=null)
{
displaybitmap(file_name);
}
else
{ //Toast.makeText("Memory Error");
//my question is how to handle two exception in UI where memory error and also
// when entered url is wrong, file not found exceptions also to handle.
}