By using java.net.URI class, how can I render an URI like this?
http://www.mysite.org/do_something?username=j.doe?firstName=John?lastName=Doe
As found by Googling java.net.URI, here's a snippet from the javadocs
A hierarchical URI is subject to further parsing according to the
syntax
[scheme:][//authority][path][?query][#fragment]
And here's a constructor from the same docs.
URI(String scheme, String authority, String path, String query, String fragment)
So I suppose
new URI(http","www.mysite.org","/do_something","username=j.doe&firstName=John&lastName=Doe","");
would do the trick.
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)
Well there are lot of discussion, post, comments and questions over internet to differentiate URI, URL and URN.
One answer on SO explain about it, but i am confused in implementation result in my code.
Q : If URI is super set of URL then how come it got this following output:
URI : /XXX/abc.do
URL : http://examplehost:8080/XXX/abc.do
When i write the below code:
System.out.println(“URI : “+ httpRequestObj.getRequestURI());
System.out.println(“URL : “+ httpRequestObj.getRequestURL());
EDIT : Could you share a detailed answer by keeping JAVA and original concept of URI,URL and URN in scope.
Regards,
Arun Kumar
java.net.URI API gives a good explanation:
A URI is a uniform resource identifier while a URL is a uniform resource locator. Hence every URL is a URI, abstractly speaking, but not every URI is a URL. This is because there is another subcategory of URIs, uniform resource names (URNs), which name resources but do not specify how to locate them. The mailto, news, and isbn URIs shown above are examples of URNs.
If URI is super set of URL then how come it got this following output ...
The definitions of URI and URL cannot be used to infer the behaviour of getRequestURI() and getRequestURL(). To understand what the methods return, you need to read the javadocs and the Servlet specification.
The meaning of those methods are what they are because the HttpRequest API has evolved over time, and that evolution has had to maintain backwards compatibility.
getRequestURI() does return a URI, and getRequestURL() does return a URL, but the URI and URL are for different things.
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"));
How can I encode URL query parameter values? I need to replace spaces with %20, accents, non-ASCII characters etc.
I tried to use URLEncoder but it also encodes / character and if I give a string encoded with URLEncoder to the URL constructor I get a MalformedURLException (no protocol).
URLEncoder has a very misleading name. It is according to the Javadocs used encode form parameters using MIME type application/x-www-form-urlencoded.
With this said it can be used to encode e.g., query parameters. For instance if a parameter looks like &/?# its encoded equivalent can be used as:
String url = "http://host.com/?key=" + URLEncoder.encode("&/?#");
Unless you have those special needs the URL javadocs suggests using new URI(..).toURL which performs URI encoding according to RFC2396.
The recommended way to manage the encoding and decoding of URLs is to use URI
The following sample
new URI("http", "host.com", "/path/", "key=| ?/#ä", "fragment").toURL();
produces the result http://host.com/path/?key=%7C%20?/%23ä#fragment. Note how characters such as ?&/ are not encoded.
For further information, see the posts HTTP URL Address Encoding in Java or how to encode URL to avoid special characters in java.
EDIT
Since your input is a string URL, using one of the parameterized constructor of URI will not help you. Neither can you use new URI(strUrl) directly since it doesn't quote URL parameters.
So at this stage we must use a trick to get what you want:
public URL parseUrl(String s) throws Exception {
URL u = new URL(s);
return new URI(
u.getProtocol(),
u.getAuthority(),
u.getPath(),
u.getQuery(),
u.getRef()).
toURL();
}
Before you can use this routine you have to sanitize your string to ensure it represents an absolute URL. I see two approaches to this:
Guessing. Prepend http:// to the string unless it's already present.
Construct the URI from a context using new URL(URL context, String spec)
So what you're saying is that you want to encode part of your URL but not the whole thing. Sounds to me like you'll have to break it up into parts, pass the ones that you want encoded through the encoder, and re-assemble it to get your whole URL.