Automatically remove Subversion unversioned files

Question

Does anybody know a way to recursively remove all files in a working copy that are not under version control? (I need this to get more reliable results in my automatic build

Answer 1

I stumbled on svn-clean on my RH5 machine. Its located at /usr/bin/svn-clean

http://svn.apache.org/repos/asf/subversion/trunk/contrib/client-side/svn-clean

Answer 2

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.

Answer 3

Since everyone else is doing it...

svn status | grep ^? | awk '{print $2}' | sed 's/^/.\//g' | xargs rm -R

Answer 4

If you are on windows command line,

for /f "tokens=2*" %i in ('svn status ^| find "?"') do del %i

Improved version:

for /f "usebackq tokens=2*" %i in (`svn status ^| findstr /r "^\?"`) do svn delete --force "%i %j"

If you use this in a batch file you need to double the %:

for /f "usebackq tokens=2*" %%i in (`svn status ^| findstr /r "^\?"`) do svn delete --force "%%i %%j"

Answer 5

Just do it on unix-shell with:

rm -rf `svn st . | grep "^?" | cut -f2-9 -d' '`

Answer 6

Can you not just do an export to a new location and build from there?