This method returns the source of the given URL.
private static String getUrlSource(String url) {
try {
URL localUrl = null;
localUrl = n
Just use
return ma.toString();
instead of
return ma;
ma.toString()
returns the string representation for your StringBuilder.
See StringBuilder#toString() for details
As Valeri Atamaniouk suggested in comments, you should also return something in the catch
block, otherwise you will get a compiler error for missing return statement
, so editing
} catch (Exception e) {
Log.e("ERR",e.getMessage());
}
to
} catch (Exception e) {
Log.e("ERR",e.getMessage());
return null; //or maybe return another string
}
Would be a good idea.
EDIT
As Esailija suggested, we have three anti-patterns in this code
} catch (Exception e) { //You should catch the specific exception
Log.e("ERR",e.getMessage()); //Don't log the exception, throw it and let the caller handle it
return null; //Don't return null if it is unnecessary
}
So i think it is better to do something like that:
private static String getUrlSource(String url) throws MalformedURLException, IOException {
URL localUrl = null;
localUrl = new URL(url);
URLConnection conn = localUrl.openConnection();
BufferedReader reader = new BufferedReader(
new InputStreamReader(conn.getInputStream()));
String line = "";
String html;
StringBuilder ma = new StringBuilder();
while ((line = reader.readLine()) != null) {
ma.append(line);
}
return ma.toString();
}
And then, when you call it:
try {
String urlSource = getUrlSource("http://www.google.com");
//process your url source
} catch (MalformedURLException ex) {
//your url is wrong, do some stuff here
} catch (IOException ex) {
//I/O operations were interrupted, do some stuff here
}
Check these links for further details about Java Anti-Patterns: