Develop Reference

URL Path Parameter Encoding

I'm struggling with path param encoding with retrofit:
http://localhost:8080/nuxeo/api/v1 is my base url.
I have this Call #GET("path/{documentPath}")
Call<Document> fetchDocumentByPath(#Path("documentPath") String docPath);
As param, I'm setting the following: default-domain/blabla
I run the query against my tomcat app and I get this answer
Response{protocol=http/1.1, code=400, message=Bad Request, url=http://localhost:8080/nuxeo/api/v1/path/default-domain%2Fblabla}
Even if I put encode = true to say "don't encode my parameter, it's already encoded", it's still encoding it.
Moreover, in retrofit, this test retrofit2.RequestBuilderTest#getWithEncodedPathParam doesn't work if we put Request request = buildRequest(Example.class, "po/ng"); with the following assertion: assertThat(request.url().toString()).isEqualTo("http://example.com/foo/bar/po/ng/");
Tomcat has restricted his URL validation for security reason: http://cve.mitre.org/cgi-bin/cvename.cgi?name=CVE-2007-0450.
So I'd like to send '/' directly in my path parameter without encoding it in %2F. How can I achieve it?
Thank you!

Since parent-2.0.0-beta4, the parameter of the annotation of #Path is now working properly.

Parsing and encoding URL before sending to Rest webservice

I am developing an App where I am sending get request to REST web service using
#RequestMapping(value = "myURLpattern", method = RequestMethod.GET, produces = "image/png")
I came across a bug if somebody enters # in the URL query String it breaks from there
http://myserver.com:8080/csg?x=1&y=2#4&z=32&p=1
So I tried to encode it via filter HTTPRequestWrapper so that # will be replaced by %23 using URLEncoder.
My problem is for encoding URL, first I need to read the URL(request.getRequestURL()).
and getURL again can't read string after #.
request.getRequestURL() will return only (http://myserver.com:8080/csg?x=1&y=2)
Is there any other way to parse the complete url and encode it before sending to REST web service?

That is not a bug. Check this:
What is the meaning of # in URL and how can i use that?
A # in the url query string is wrong. You should encode it on client side, before sending it to your server. See this one: How to pass '#' in query string
It is in asp, but you should find the java equivalent.
Something to get you started can be this one Java URL encoding of query string parameters

Invoking a 'REST' service which have query parameters in the URL

I have to invoke a GET on a service which returns text/xml.
The endpoint is something like this:
http://service.com/rest.asp?param1=34&param2=88&param3=foo
When I hit this url directly on a browser (or some UI tool), all's good. I get a response.
Now, I am trying to use CXF WebClient to fetch the result using a piece of code like this:
String path = "rest.asp?param1=34&param2=88&param3=foo";
webClient.path(path)
.type(MediaType.APPLICATION_JSON)
.accept(MediaType.TEXT_XML_TYPE)
.get(Response.class);
I was debugging the code and found that the request being sent was url encoded which appears something like this:
http://service.com/rest.asp%3Fparam1=34%26param2=88%26param3=foo
Now, the problem is the server doesn't seem to understand this request with encoded stuff. It throws a 404. Hitting this encoded url on the browser also results in a 404.
What should I do to be able to get a response successfully (or not let the WebClient encode the url)?

Specify the parameters using the query method:
String path = "rest.asp";
webClient.path(path)
.type(MediaType.APPLICATION_JSON)
.accept(MediaType.TEXT_XML_TYPE)
.query("param1","34")
.query("param2","88")
.query("param3","foo")
.get(Response.class);

You will need to encode your URL. You can do it with the URLEncoder class as shown below:
Please replace your line
String path = "rest.asp?param1=34&param2=88&param3=foo";
with
String path = URLEncoder.encode("rest.asp?param1=34&param2=88&param3=foo");

How to encode a URL that I want to pass as a query string

I'm trying to send a URL as paramter of a query string like this example:
http://localhost.com/myapp.jsp?pathToFileURL=http://192.168.0.1/my_file.pdf
What I did is I used encode URL to encode the path before sending it to the server, problem is im getting a "400 Invalid URI: noSlash" because of this.
From what I read the problem is the tomcat security and that I should add a parameter to the tomcat startup
-Dorg.apache.tomcat.util.buf.UDecoder.ALLOW_ENCODED_SLASH=true
But I can't modify the parameters of the tomcat, so is it possible to do it other way?
Thanks

You can do URLSafebase64 encoding at the client side and URLSafebase64 decoding at the server side.
Check URLEncoder class for more details:
http://docs.oracle.com/javase/1.5.0/docs/api/java/net/URLEncoder.html
You can test manually before coding using any of the online URL Encoder/Decoder. Just google for "URL Encoder/Decoder"

Complete stab in the dark but you could try escaping the slashes with backslashes or you could try replacing them with %2F which is the URL encoded version of forward slash.
Hope this helps.

Base64 the URL then on the receiving end base64 decode to get the original URL without any alteration

Safe Data serialization for Plain HTTP GET & POST communication

I'm using the client's browser to submit HTTP request.
For report generation the securityToken is submitted as POST, for report download the same token needs to be submitted by the user browser, this time using GET.
What encoding would you recommend for the securityToken which actually represents encrypted data.
I've tried BASE64 but this fails because the standard can include the "+" character which gets translated in HTTP GET to ' ' (blank space).
Then I tried URL Encoding, but this fails because for HTTP POST stuff such as %3d are transmitted without translation but when browser does HTTP GET with the data %3d is converted to '='.
What encoding would you recommend, to allow safe transmission over HTTP POST & GET without data being misinterpreted.
The environment is Java, Tomcat.
Thank you,
Maxim.

Hex string.
Apache commons-codec has a Hex class that provides this functionality.
It will look like this:
http://youraddress.com/context/servlet?param=ac7432be432b21

Well, you can keep the Base64 and use this solution:
Code for decoding/encoding a modified base64 URL