Android URL to Uri

Question

I have to change URL to URI, please guide me how can I change it.

I have tried a lot, but not getting the solution. Any help is appreciated. I can also attach code snippet if required.

see this link http://stackoverflow.com/questions/2593214/android-howto-parse-url-string-with-spaces-to-uri-object — , Mar 12 '12 at 07:04

Samir Mangroliya · Answer 1 · 2020-06-06T10:52:02.837

125

We can parse any URL using Uri.parse(String) method.

Code Snippet :

Uri uri =  Uri.parse( "http://www.stackoverflow.com" );

edited Jun 06 '20 at 10:52

answered Mar 12 '12 at 07:03

Samir Mangroliya

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5

Well that would be String to Uri, not URL to Uri. – tir38 Oct 31 '14 at 20:10
6

should `java.net.URI` be used or `android.net.Uri` ? Is there a preference ? – Someone Somewhere Jan 14 '15 at 19:33

score 35 · Answer 2 · edited Jan 11 '15 at 13:20

35

final String myUrlStr = "xyz";
URL url;
Uri uri;
try {
    url = new URL(myUrlStr);
    uri = Uri.parse( url.toURI().toString() );
} catch (MalformedURLException e1) {
    e1.printStackTrace();
} catch (URISyntaxException e) {
    e.printStackTrace();
}

edited Jan 11 '15 at 13:20

JJD

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answered Mar 12 '12 at 07:10

Sparky

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At first glance that answer looks convoluted. Why convert to string and then parse? The reason is that `toURI()` returns a `java.net.URI`. So we need to map to string and then parse into `android.net.Uri`. Good answer – tir38 Oct 31 '14 at 20:13

score 1 · Answer 3 · answered May 20 '13 at 05:14

This is my 2 Cents , it could help someone else.

The standard URL Parsing might not be good idea if you have variables passed to the URL as Query

String your_url_which_is_not_going_to_change = "http://stackoverflow.com/?q="
String your_query_string = "some other stuff"
String query = null;

try {
  query = URLEncoder.encode(your_query_string, "utf-8");
} catch (UnsupportedEncodingException e) {
  // TODO Auto-generated catch block
  e.printStackTrace();
}

String url = your_url_which_is_not_going_to_change + query;

URL input = new URL(url);
etc... .

Hope it helps.

H.

he is asking to parse URL to URI however you are only parsing the URL — IIRed-DeathII, Jun 14 '17 at 14:11

score 0 · Answer 4 · answered Sep 26 '21 at 11:33

0

var url = "http://10.0.2.2:8080/(mymobileapp)/login.php";

var response = await http.post(Uri.parse(url), body: data);

answered Sep 26 '21 at 11:33

Walied Abdulaziem

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Android URL to Uri

4 Answers4