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.
Related
I have a request as follows:
localhost:8000/location/:01
My code takes as input an HttpContext request.
func(HttpExchange r) {
String area_path = r.getRequestURI(); // Equals string "/location/"
}
How do I parse an HttpExchange correctly so I can pull out the "01" from this path and store it as a variable?
That (localhost:8000/location/:01) is not a valid URL or URI
A plain colon character is not legal in the path of a URL or URI. If you want to put a colon in the path, it must be percent-encoded. Furthermore, if this was a URL, it would start with a protocol; e.g. http:.
Now ... it is unclear what the HTTP stack you are using will do with a syntactically incorrect URL / URI, but it could simply be ignoring the colon and the characters after it.
Your code looks a bit odd too. You have tagged the question as [java]. But the code looks like JavaScript rather than Java; i.e. func is a Javascript keyword. But it also looks like you are using the (deprecated) com.sun.net.httpserver.HttpExchange Java class. I don't know what to make of that ...
My advice:
Don't use a colon character in the URL path.
If you must do it, then percent-encode the colon it.
If you cannot encode it properly, then you may need to find and use a different framework for your HTTP request handling. One that will accept and handle a malformed URL / URI in the way that you want. (Good luck finding one!)
Unfortunately, the details in your question are too sketchy to give more detailed advice.
I was wondering if I can hardcore the file:// prefix into one of my functions in android.
The function is supposed to determine whether or not the given link points to an external resource on the web, or an internal resource inside the phone itself
public Uri generate_image_uri(String link)
{
// link can be "1DCHiI2.jpg"
// link can be "file://smiley_face.jpg"
if (!link.startsWith("file://")
return Uri.parse("https://i.imgur.com/" + link);
else
return Uri.parse(link);
}
Is this advisable? Or is there a more "fault tolerant" way of getting file://? maybe some function like getProperFilePrefixForThisAndroidVersion();?
In order to clarify my question:
given the following code
(new File(getFilesDir(), "hello_world.jpg")).toString();
Is it safe to assume within reasonable probability that the resulting string will always start with file:// in all current and future Android versions?
Given:
(new File(getFilesDir(), "hello_world.jpg")).toString();
is it safe to assume within reasonable probability that the resulting string will always start with file:// in all current and future Android versions?
No.
According to the javadoc for File on Android, File.toString returns:
"... the pathname string of this abstract pathname. This is just the string returned by the getPath() method."
Not a "file://" URL.
If you want to get a properly formed "file://" URL, do this:
new File(...).toURI().toString()
Now, technically the protocol for a URL (i.e. "file") is case insensitive:
Is the protocol name in URLs case sensitive?
Which means that "FILE://" or "File://" etc are technically valid alternatives.
However, the probability that above expression would ever emit anything other than the lower-case form of the protocol is (um) vanishingly small1.
1 - It would entail monumentally stupid decision making by a number of people. And they are NOT stupid people.
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 can't find any explanation as to what exactly the "scheme-specific part" of a URI is.
From wikipedia :
All URIs and absolute URI references are formed with a scheme name,
followed by a colon character (":"), and the remainder of the URI
called (in the outdated RFCs 1738 and 2396, but not the current STD
66/RFC 3986) the scheme-specific part.
The scheme-specific-part is what you have after the :.
Example :
http://stackoverflow.com/questions/24077453/
scheme : scheme-specific-part
Each URI begins with a scheme name that refers to a specification for assigning identifiers within that scheme. As such, the URI syntax is a federated and extensible naming system wherein each scheme's specification may further restrict the syntax and semantics of identifiers using that scheme.
See this section of the URI rfc https://www.rfc-editor.org/rfc/rfc3986#section-3.1
Scheme specific means just to simple define which Protocol is used by the Url like
HTTP or HTTPS .
So simply add these in URL to work fine
Scheme Specific
http://localhost:8080/api/notes
Without Scheme
localhost:8080/api/notes
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.