URI.getPath（）在“＃”之后不返回任何内容

Question

I have a URI with path like this :

ftp://test:test@someftp/ready123/users/abc/#0#.

But the method getPath() on the URI returns this:

/ready123/users/abc/. 

I need the whole path to be returned like this :

/ready123/users/abc/#0#

... So that I can change the working directory(CWD) to the folder #0#. The code is in a generic method and is jarred up which gets used by many other applications. So I have to be very careful when I make changes. I believe anything after # is considered as a fragment, but in this case it is actually the name of a folder.

How do I get the path /ready123/users/abc/#0# from the URI object ?

Answer 1

From the JavaDoc of URI class:

URI syntax and components At the highest level a URI reference (hereinafter simply "URI") in string form has the syntax

 [scheme:]scheme-specific-part[#fragment] 

where square brackets [...] delineate optional components and the characters : and # stand for themselves.

As such if you want to retrieve everything after the first # you need to use URI.getFragment()

Answer 2

You need to replace the hash character with %23

 URI uir = new URI("ftp://test:test@someftp/ready123/users/abc/%230%23");
 System.out.println(uir.getPath()); // returns '/ready123/users/abc/#0#'