We Keep Coding

How do I parse an HttpExchange request after an ending slash?

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.

Jersey Post request - How to perform a file upload with an unknown number of additional parameters?

I asked something like this previously, but upon re-reading my original post, it was not easy to understand what I was really asking. I have the following situation. We have (or at least I'm trying to get working) a custom file upload procedure that will take in the file, a set number of 'known' metadata values (and they will always be there), as well as potentially an unknown number of additional metadata values. The service that exists currently uses the Jersey framework (1.16)
I currently have both client and server code that handles dealing with the file upload portion and the known metadata values (server code below)
#POST
#Path("asset/{obfuscatedValue0}/")
#Consumes(MediaType.MULTIPART_FORM_DATA)
public UUID uploadBlob(#PathParam("obfuscatedValue0") Integer obfuscatedValue0,
#FormDataParam("obfuscatedValue1") String obfuscatedValue1,
#FormDataParam("obfuscatedValue2") String obfuscatedValue2,
#FormDataParam("obfuscatedValue3") String obfuscatedValue3,
#FormDataParam("obfuscatedValue4") String obfuscatedValue4,
#FormDataParam("obfuscatedValue5") String obfuscatedValue5,
#FormDataParam("file") InputStream uploadedInputStream) {
.....
}
...and excerpt of client code:
Builder requestBuilder = _storageService
.path("asset")
.path(obfuscatedValue0.toString())
.type(MediaType.MULTIPART_FORM_DATA)
.accept(MediaType.APPLICATION_JSON);
FormDataMultiPart part = new FormDataMultiPart()
.field("file", is, MediaType.TEXT_PLAIN_TYPE) // 'is' is an inputstream from earlier in code.
.field("obfuscatedValue1", obfuscatedValue1)
.field("obfuscatedValue2", obfuscatedValue2)
.field("obfuscatedValue3", obfuscatedValue3)
.field("obfuscatedValue4", obfuscatedValue4)
.field("obfuscatedValue5", obfuscatedValue5);
storedAsset = requestBuilder.post(UUID.class, part);
However, I need to pass a map of additional parameters that will have an unknown number of values/names. From what I've seen, there is no easy way to do this using the FormDataParam annotation like my previous example.
Based upon various internet searches related to Jersey file uploads, I've attempted to convert it to use MultivaluedMap with the content type set to "application/x-www-form-urlencoded" so it resembles this:
#POST
#Path("asset/{value}/")
#Consumes("application/x-www-form-urlencoded")
public UUID uploadBlob(#PathParam(value), MultivaluedMap<String,String> formParams) {
....
}
It's my understanding that MultivaluedMap is intended to obtain a general map of form parameters (and as such, cannot play nicely together in the same method bearing #FormDataParam annotations.) If I can pass all this information from the Client inside some sort of map, I think I can figure out how to handle parsing the map to grab and 'doMagic()' on the data to get what I want done; I don't think I'll have a problem there.
What I AM fairly confused about is how to format the request client-side code when using this second method within the jersey framework. Can anyone provide some guidance for the situation, or some suggestions on how to proceed? I'm considering trying the solution proposed here and developing a custom xml adapter to deal with this situation, and sending xml instead of multipart-form-data but I'm still confused how this would interact with the InputStream value that will need to be passed. It appears the examples with MultivaluedMap that I've seen only deal with String data.

Special characters in XML response received from server

In my Scala code, I am fetching a response from a server using the getInputStream method of HttpUrlConnection class. The response is XML data. However the data contains HTML entities like & and '.
Is there a way I can replace these characters with their text equivalent so that I can parse the XML properly?

It's necessary to encode those entities in xml so they don't interfere with its syntax. The <(<) and > (>) entities make this more obvious. It would be impossible to parse XML whose content was littered with < and > symbols.
Scala's scala.xml package should give you the tools you need to parse your xml. Here's some guidance from the library's author.

Tomcat Servlet FileUpload and UTF-8

I am using commons FileUpload to get client headers. I am trying to use the header Strings to write txt content like:
StringA; StringB; "ciryllic word"
The server localization is ru_RU.utf-8 the client is cp1251... The result is I always get on my client (desktop app)
StringA; StringB; ?????
instead of cyrillic characters in my server txt files lines. If I open txt with my IE 8 and watch the encoding the cyrillic content can be readable with utf-8 encoding only :( So my question is what should me do to make my servlet write ru_RU.utf-8 acceptable strings. I am testing it on windows so I need to know it for sure
Please help me to understand
Any useful comment is appreciated

ServletOutputStream out =response.getOutputStream();
out.write((YOUrfileasstring).getBytes("UTF-8"));

I might be misunderstanding the question, but to answer this part -
So my question is what should me do to make my servlet write ru_RU.utf-8 acceptable strings.
You should use either of two methods provided by the HttpServletResponse class. Those are:
setContentType(), e.g.
response.setContentType("text/html; charset=UTF-8");
or,
setCharacterEncoding(), e.g.
response.setCharacterEncoding("utf-8");
There are other examples for those methods and information on when to use them all over the place.

Getting binary file and text information

Currently, I have a servlet act as web service.
When I pass in parameters using POST, it will return me an executable binary file (application/octet-stream). However, beside binary file, I would also like to get additional information (in text format) about this binary file.
Is it possible to achieve this by using only single POST request? But, how is it possible, to switch from application/octet-stream to text/plain within single POST response?

It is not possible to change the MIME type within a single response.
However, i think t is possible to put your additional information into the response header using the HttpServletResponse.addHeader method.

You could return a multipart MIME response (multipart/mixed; boundary=XXX instead of application/octet-stream) with the binary part encoded in Base64.
I'm not sure if the JavaMail API can be used to construct the content, but it's worth a look.

Within a single POST it isn't possible. But two ideas:
Show your text file as another web page that starts the download of your executable
Bundle both files into a zip archive