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I have a java program that is supposed to read a file from a URL (the URL location is a virtual directory under an IIS web site; below, and in my initial testing, I'm treating it like any other file system location). Unfortunately the path to all the files that need to be read includes a pound sign (#) in one of the directory names, and there’s nothing I can do to change that. The program works beautifully when (as a test) I point it to a location that doesn't have that pound sign in the path.

I started by creating a URL from a string passed to the program. For a file path like /Documents/#2012/09/11 (where Documents is a Windows share), I could get the program to process successfully if I passed it a path like this on my command line:

file://serverIPaddress/Documents/\%232012/09/07/16/DOC4671179.DOC

That is, with the pound sign manually encoded as %23, and a back slash escaping the % of the %23.

There was just one line to get that URL: 

URL url = new URL(filePath); // filePath is passed in

But the program isn’t going to be spoon-fed an encoded path like that, so I had to figure out how to encode the pound sign programmatically. Going on the good advice found at how to encode URL to avoid special characters in java, I created a URI using a multi-argument constructor (I broke up the parameter I had been passing to the program into three separate parameters to accommodate that change). Here’s what that looked like:

URI uri = new URI(protocol, host, filePath, null); // all values are passed in

That encoded the pound sign properly; my URI was:

file://serverIPaddress/Documents/%232012/09/07/16/DOC4671179.DOC

But without the backslash in front of the %23, the program came back with Connection refused, presumably because it’s misinterpreting the path without the benefit of that backslash.

So I thought, ok, I’ll add the backslash myself. I created the same URI, extracted its rawPath, and with a bit of string manipulation, put a backslash in front of the %23. I then created a new URI using that new string:

URI uri = new URI(protocol, host, filePath, null); // all values are passed in
String rawPath = uri.getRawPath();
int pctPos = rawPath.indexOf("%");
String escaped = new String("\\");
String firstPart = rawPath.substring(0,pctPos);
String secondPart = rawPath.substring(pctPos);
String newPath = firstPart + escaped + secondPart;
URI uri2 = new URI(protocol, host, newPath, null);

However, predictably, that gave me a URI like this:

file://<serverIPaddress>/Documents/%5C%25232012/09/07/16/DOC4671179.DOC

with both the backslash and the % encoded. Makes sense, but still doesn’t work at execution time.

The URL API says:

The URL class does not itself encode or decode any URL components according to the escaping mechanism defined in RFC2396. It is the responsibility of the caller to encode any fields, which need to be escaped prior to calling URL

So I thought, ok, instead of creating a second URI, I’ll create a URL from that new string I generated in the last try:

URI uri = new URI(protocol, host, filePath, null); // all values are passed in
String rawPath = uri.getRawPath();
int pctPos = rawPath.indexOf("%");
String escaped = new String("\\");
String firstPart = rawPath.substring(0,pctPos);
String secondPart = rawPath.substring(pctPos);
String newPath = firstPart + escaped + secondPart;
URL url = new URL(protocol + "://" + host + newPath);

But in that approach, even though my new path looked good as:

/Documents/\%232012/09/07/16/DOC4671179.DOC

the resulting URL comes back as:

file://serverIPAddress/Documents//%232012/09/07/16/DOC4671179.DOC

with an extra forward slash in front of the %23 instead of a backslash.

And with that I’ve run out of ideas.

  • What makes the back slash in this last approach turn into a forward slash in the URL?

  • What can I do go get the URI/URL I need?

  • Or maybe I should ask: why does the program need the % in the %23 to be escaped in the first place, if that %23 is part of a legitimate URI or URL, and is there something I can do about that instead?

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user1664369
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  • What are you passing the URL to that gives "connection refused"? – nneonneo Sep 12 '12 at 01:57
  • Some info in this related ServerFault question: http://serverfault.com/q/257680 – barrowc Sep 12 '12 at 02:51
  • nneonneo - Not sure I'm understanding your question. The program is opening a connection to the URL, getting an InputStream, then an InputStreamReader, then a BufferedReader, and then reading from the file. The connection refused error comes when trying to get the InputStream. URLConnection urlConn = url.openConnection(); urlConn.setDoInput(true); InputStream is = urlConn.getInputStream(); InputStreamReader isr = new InputStreamReader(is); br = new BufferedReader (isr); – user1664369 Sep 12 '12 at 03:42

1 Answers1

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not sure why "\" is required. it depends on server code. actually "\" is not a legal char in URLs, it should be encoded as %5C

URI class is pretty messy. It might silently change "\" to "/" for file URLs.

Try this instead:

    String filePath = "/Documents/#2012/09/11";
    filePath = filePath.replace("#", "\\#");
    URI uri = new URI("file", "serverAddress", filePath, null);

The "#" will be changed to "%5C%23". see if it works.

irreputable
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  • Thanks for the suggestion. As predicted the `replace` gives the URI `file://serverAddress/Documents/%5C%232012/09/11/filename`, but that URI also results in an "IOException: connection refused" error when trying to get the InputStream (after creating a URL from the URI, opening a connection to the URL, etc). – user1664369 Sep 12 '12 at 05:39
  • how about `new URL("file://"+serverAddress+filePath.replace("#", "\\%23")` – irreputable Sep 12 '12 at 15:48
  • Good idea. But that also comes back with a URL of `file://172.28.60.68/Documents//%232012/09/11/DOC4671179.DOC` - a forward slash instead of the requested back slash. There’s something about that class that doesn’t like the back slash. If the File class will accept a UNC path in its string constructor maybe that will work. I’m being pulled away by production issues but I’ll try it when I get a moment. – user1664369 Sep 12 '12 at 16:43
  • I gave a copy of the initial version of my java program (the one that creates a URL from a string) to a co-worker who is building the process that will call the program. He reports that he can call the program successfully by encoding the # to %23 - no backslash. So I guess we can leave it at that. For testing from a command line we can call it with \%23, and for calling it programmatically we'll call it with normal encoding. I would like to know *why* the backslash is necessary from a command line, but I guess we can chalk it up to the command line processor. Thank you for all the help! – user1664369 Sep 12 '12 at 21:43
  • @user1664369 - by 'command line' I expect you mean as a parameter in a command executed by typing into a shell. In that case you should be aware that `%` is probably a special character. In `bash`, for example, I believe it is _remainder on division_, and it is used as an operator in variable patterns, too. So it needs to be escaped _for the shell_. If you put the entire URI in single quotes in the shell command, you shouldn't need the backslash. – Steve Powell Sep 11 '13 at 13:29