I have a string, 'songchoice'. I want it to become a 'Uri' so I can use with MediaPlayer.create(context, Uri)
How can I convert songchoice to the Uri?
Uri myUri = Uri.parse("http://www.google.com");
Here's the doc http://developer.android.com/reference/android/net/Uri.html#parse%28java.lang.String%29
Uri.parse(STRING);
See doc:
String: an RFC 2396-compliant, encoded URI
Url must be canonicalized before using, like this:
Uri.parse(Uri.decode(STRING));
The String I had to convert to a URI was a local file path, but without the file:// prefix. So even
Uri.parse(Uri.decode(STRING));
resulted in FileNotFoundException: No content provider (see also this question).
I got a valid URI by first creating a File object i.e.:
Uri myUri = Uri.fromFile(new File(STRING));
Related
I have a URI like this:
java.net.URI location = UriBuilder.fromPath("../#/Login").queryParam("token", token).build();
and I am sending it as response: return Response.seeOther(location).build()
However, in the above URI, # is getting encoded to %23/. How do I create a URI with out encoding the hash #. According to official document, a fragment() method must be used to keep unencoded.
URI templates are allowed in most components of a URI but their value
is restricted to a particular component. E.g.
UriBuilder.fromPath("{arg1}").build("foo#bar"); would result in
encoding of the '#' such that the resulting URI is "foo%23bar". To
create a URI "foo#bar" use
UriBuilder.fromPath("{arg1}").fragment("{arg2}").build("foo", "bar") instead.
Looking at the example from docs, I am not sure how to apply it in my case.
The final URI should look like this:
http://localhost:7070/RTH_Sample14/#Login?token=eyJhbGciOiJSUzI1NiJ9.eyJpc3MiOiJodHRwczpcL1wvcnRoLmNvbSIsInN1YiI6IlJUSCIsInJvbGUiOiJVU0VSIiwiZXhwIjoxNDU2Mzk4MTk1LCJlbWFpbCI6Imtpcml0aS5rOTk5QGdtYWlsLmNvbSJ9.H3d-8sy1N-VwP5VvFl1q3nhltA-htPI4ilKXuuLhprxMfIx2AmZZqfVRUPR_tTovDEbD8Gd1alIXQBA-qxPBcxR9VHLsGmTIWUAbxbyrtHMzlU51nzuhb7-jXQUVIcL3OLu9Gcssr2oRq9jTHWV2YO7eRfPmHHmxzdERtgtp348
To construct the URI with fragment use
UriBuilder.fromPath("http://localhost:7070/RTH_Sample14/").fragment("Login").build()
This results in the URI string
http://localhost:7070/RTH_Sample14/#Login
But if you also add query parameters
UriBuilder.fromPath("http://localhost:7070/RTH_Sample14/").fragment("Login")
.queryParam("token", "t").build()
then the UriBuilder always inserts the query params before the fragment:
http://localhost:7070/RTH_Sample14/?token=t#Login
which simply complies to the URL syntax.
Instead of all the hassle of redirecting without encoding the hash value. I changed my code into the following:
java.net.URI location = new java.net.URI("../#/Login?token=" + token);
So the query param above is token appended to URI location. In front-end I am using angular's location.search().token to get capture the query param.
This worked for me. Looking for better answers though. Thanks
I need a solution to convert file path to EMF URI, not a Java URI.
I tried with this:
org.eclipse.emf.common.util.URI ur = org.eclipse.emf.common.util.URI.createURI(URI.createURI(file.getPath()).toString());
...but I get this exception:
java.net.MalformedURLException: unknown protocol: c appears .
Is there another solution?
In EMF, URI classe comes with a lot of static methods to help you create your URI. In your specific case, try URI.createFileURI(...) instead of URI.createURI(...)
URI fileURI = URI.createFileURI(file.getAbsolutePath());
Look here for details about the method: http://download.eclipse.org/modeling/emf/emf/javadoc/2.4.3/org/eclipse/emf/common/util/URI.html#createFileURI(java.lang.String)
I am trying to read a file using URL in java.
FileHelper.read(new File(getClass.getResource("TextFile.rtf")))
I am really confused with the below exception
error: overloaded method constructor File with alternatives:
(java.net.URI)java.io.File <and>
(java.lang.String)java.io.File
cannot be applied to (java.net.URL)
Any idea or suggestion how can I resolve this exception.
Thanks !!!
Try converting the URL to its URI equivalent:
FileHelper.read(new File(getClass.getResource("TextFile.rtf").toURI()))
See URL.toURI() for more information.
All you need to do is to get URI from the URL.
URL url = getClass.getResource("TextFile.rtf");
URI uri = url.toURI();
FileHelper.read(new File(uri))
Given an URL such as this one,
http://www.example.com/some directory/some file
how do you encode this URL? Browsers automatically encode it. In Java I couldn't find a ready made function. I suspect there should be such a function because this is generally needed.
When I try to use the URI class using the constructor with single String, and parse components of the URL, such as authority, path, etc, it gives error because it expects an encoded URL.
Do you know a ready made function that will produce, for example in this case:
http://www.example.com/some%20directory/some%20file
Try this:
final URL url = new URL("http://www.example.com/some directory/some file");
final URI uri = new URI(url.getProtocol(), url.getHost(), url.getPath(), null);
System.out.println(uri.toASCIIString());
I've got a string:
public://imageifarm/3600.jpg
How can I extract the
imageifarm/3600.jpg
Part out using android?
What I've tried so far:
URL drupalQuestionNodeImageURI = new URL("public://imageifarm/3600.jpg");
Log.d("TAG", drupalQuestionNodeImageURI.getPath());
but it throws this exception:
09-16 17:24:39.992: W/System.err(3763): java.net.MalformedURLException: Unknown protocol: public
How can I solve this?
I know I can use regular expressions but that seems to defeat the purpose of URL(URI) in this case.
You should use android.net.Uri
Uri mUri = Uri.parse(public://imageifarm/3600.jpg);
String extract = mUri.getEncodedSchemeSpecificPart();
Use java.net.URI, not java.net.URL.
If you want have to use URL class (when you image sits on Internet) you have to provide valid URL (that begins from valid URL prefix, like http://, https:// etc). In you case you should use Uri class. Uri object can point on files in your local file system. For example:
Uri.fromFile(new File("public://imageifarm/3600.jpg"));