How to create a android.net.Uri from a java.net.URL?

Question

I would like to use intent.setData(Uri uri) to pass data obtained from a URL. In order to do this, I need to be able to create a Uri from a URL (or from a byte[] I read from the URL, or a ByteArrayInputStream I create from the byte[], etc). However, I cannot figure out how this is supposed to be done.

So, is there anyway to create a Uri from data obtained from a URL without first writing the data to a local file?

Buhake Sindi · Answer 1 · 2011-11-23T10:03:01.113

42

Use URL.toURI() (Android doc) method.

Example:

URL url = new URL("http://www.google.com"); //Some instantiated URL object
URI uri = url.toURI();

Make sure to handle relevant exception, such as URISyntaxException.

edited Nov 23 '11 at 10:03

answered Nov 23 '11 at 09:57

Buhake Sindi

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toURI returns "URI" from net package, setData method needs "Uri" – Strassenrenner Dec 28 '14 at 19:00
This does not take into account the special characters that a parameter might have. – Farrukh Chishti Sep 05 '16 at 12:03
@QualtarDemix that is not what was asked by the OP. I provided a solution based on this requirement. – Buhake Sindi Sep 05 '16 at 13:08
@BuhakeSindi: Well that is implicit. The idea is to get the best possible solution. Even if that wasn't explicitly asked. – Farrukh Chishti Sep 07 '16 at 10:03
Suppose **http** is not part of the url string? – IgorGanapolsky May 15 '17 at 18:15

score 13 · Answer 2 · edited Nov 23 '11 at 09:58

I think your answer can be found from here..

Uri.Builder.build() works quite well with normal URLs, but it fails with port number support.

The easiest way that I discovered to make it support port numbers was to make it parse a given URL first then work with it.

Uri.Builder b = Uri.parse("http://www.yoursite.com:12345").buildUpon();

b.path("/path/to/something/");
b.appendQueryParameter("arg1", String.valueOf(42));

if (username != "") {
  b.appendQueryParameter("username", username);
}

String url = b.build().toString();

Source : http://twigstechtips.blogspot.com/2011/01/android-create-url-using.html

Also, instead of hardcoding "http://...", you could use `yourUrlVariable.toString()` instead. — dmon, Nov 23 '11 at 16:30

score 11 · Answer 3 · edited May 23 '17 at 12:26

11

From How to create a Uri from a URL?

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

edited May 23 '17 at 12:26

Community

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1

answered Apr 15 '16 at 18:30

NathanH

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7

score 4 · Answer 4 · edited May 23 '17 at 12:03

Note that in Android, Uri's are different from Java URI's. Here's how to avoid using hardcoded strings, and at the same time, create a Uri with just the path portion of the http URL string encoded to conform to RFC2396:

Sample Url String:

String thisUrl = "http://lapi.transitchicago.com/api/1.0/ttarrivals.aspx?key=[redacted]&mapid=value"

method:

private Uri.Builder builder;
public Uri getUriFromUrl(String thisUrl) {
    URL url = new URL(thisUrl);
    builder =  new Uri.Builder()
                            .scheme(url.getProtocol())
                            .authority(url.getAuthority())
                            .appendPath(url.getPath());
    return builder.build();
}

To handle query strings you will need to parse the url.getQuery() as described here and then feed that into builder. appendQueryParameter().

score 2 · Answer 5 · answered Nov 23 '11 at 09:59

2

try {
    uri = new URI(url.toString());
} catch (URISyntaxException e) {
}

answered Nov 23 '11 at 09:59

1

Cannot instantiate the type Uri – Prasad Sep 27 '14 at 10:13

score 0 · Answer 6 · answered Nov 23 '11 at 10:18

URI uri = null;
URL url = null;

// Create a URI
try {
    uri = new URI("www.abc.com");
} catch (URISyntaxException e) {
}

// Convert an absolute URI to a URL
try {
    url = uri.toURL();
} catch (IllegalArgumentException e) {
    // URI was not absolute
} catch (MalformedURLException e) {
}

// Convert a URL to a URI
try {
    uri = new URI(url.toString());
} catch (URISyntaxException e) {
}

How to create a android.net.Uri from a java.net.URL?

6 Answers6

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