Here is my code to fetch image from an
url and then display it into imageview in android (I use android 2.2.3)
In this tutorial, I get image from this
link https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEizHvU0XODfjwRifjEZqSlbL7tLlgLVsEjUmgQnLIdtud9kr_Fm8NADiZeF-ajC5-7Um8za4iMM6mXwRAzHPGhyphenhyphenMIc_LCekHNtrDkouUI2SXospNb8DjkEAWDkPUqDBl8TIitJZ_MuTZ3c/h120/images.png
- main.xml layout:
public static String src_link = "https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEizHvU0XODfjwRifjEZqSlbL7tLlgLVsEjUmgQnLIdtud9kr_Fm8NADiZeF-ajC5-7Um8za4iMM6mXwRAzHPGhyphenhyphenMIc_LCekHNtrDkouUI2SXospNb8DjkEAWDkPUqDBl8TIitJZ_MuTZ3c/h120/images.png";
public ImageView imgView;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.main);
imgView = (ImageView)findViewById(R.id.imageView1);
imgView.setImageBitmap(fetchImage(src_link));
}
private Bitmap fetchImage(String urlstr) {
Bitmap img = null;
try {
URL url;
url = new URL(urlstr);
HttpURLConnection c = (HttpURLConnection) url.openConnection();
c.setDoInput(true);
c.connect();
InputStream is = c.getInputStream();
img = BitmapFactory.decodeStream(is);
return img;
} catch (MalformedURLException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
return img;
}
- Finally, add permission in Manifest.xml:
Yep! This is end of this topic. Hope you enjoy it! :)
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