Automatically remove Subversion unversioned files

I ran across this page while looking to do the same thing, though not for an automated build.

After a bit more looking I discovered the 'Extended Context Menu' in TortoiseSVN. Hold down the shift key and right click on the working copy. There are now additional options under the TortoiseSVN menu including 'Delete unversioned items...'.

Though perhaps not applicable for this specific question (i.e. within the context of an automated build), I thought it might be helpful for others looking to do the same thing.

this works for me in bash:

 svn status | egrep '^\?' | cut -c8- | xargs rm

Seth Reno's is better:

svn status | grep ^\? | cut -c9- | xargs -d \\n rm -r 

It handles unversioned folders and spaces in filenames

As per comments below, this only works on files that subversion doesn't know about (status=?). Anything that subversion does know about (including Ignored files/folders) will not be deleted.

If you are using subversion 1.9 or greater you can simply use the svn cleanup command with --remove-unversioned and --remove-ignored options

Edit:

Subversion 1.9.0 introduced an option to do this:

svn cleanup --remove-unversioned

Before that, I use this python script to do that:

import os
import re

def removeall(path):
    if not os.path.isdir(path):
        os.remove(path)
        return
    files=os.listdir(path)
    for x in files:
        fullpath=os.path.join(path, x)
        if os.path.isfile(fullpath):
            os.remove(fullpath)
        elif os.path.isdir(fullpath):
            removeall(fullpath)
    os.rmdir(path)

unversionedRex = re.compile('^ ?[\?ID] *[1-9 ]*[a-zA-Z]* +(.*)')
for l in  os.popen('svn status --no-ignore -v').readlines():
    match = unversionedRex.match(l)
    if match: removeall(match.group(1))

It seems to do the job pretty well.