On Thu, 2006-04-27 at 15:47 -0400, Bill Marshall wrote:
> Hi All:
>
> I am having a problem setting up ViewVC on a Windows XP Home machine
> running Apache 2.0. I am getting the following error when attempting
> to access a folder with files in it:
>
> An Exception Has Occurred
>
> Python Traceback
>
> Traceback (most recent call last):
>
> File "C:\Program Files\viewvc-1.0.0-rc1\lib\viewvc.py", line 3609,
> in main
>
> request.run_viewvc()
>
> File "C:\Program Files\viewvc-1.0.0-rc1\lib\viewvc.py", line 386, in
> run_viewvc
>
> self.view_func(self)
>
> File "C:\Program Files\viewvc-1.0.0-rc1\lib\viewvc.py", line 1539,
> in view_directory
>
> request.repos.dirlogs(request.path_parts, request.pathrev,
> file_data, options)
>
> File "C:\Program Files\viewvc-1.0.0-rc1\lib\vclib\bincvs
> \__init__.py", line 195, in dirlogs
>
> alltags = _get_logs(self, dirpath, entries, rev, subdirs)
>
> File "C:\Program Files\viewvc-1.0.0-rc1\lib\vclib\bincvs
> \__init__.py", line 880, in _get_logs
>
> rlog = repos.rcs_popen('rlog', args, 'rt')
>
> File "C:\Program Files\viewvc-1.0.0-rc1\lib\vclib\bincvs
> \__init__.py", line 255, in rcs_popen
>
> return popen.popen(cmd, args, mode, capture_err)
>
> File "C:\Program Files\viewvc-1.0.0-rc1\lib\popen.py", line 71, in
> popen
>
> phandle, pid, thandle, tid = win32popen.CreateProcess(command,
> hStdIn, hStdOut, hStdErr)
>
> File "C:\Program Files\viewvc-1.0.0-rc1\lib\win32popen.py", line 74,
> in CreateProcess
>
> si # startupinfo
>
> error: (5, 'CreateProcess', 'Access is denied.')
>
>
>
> I have guessed that the python script does not have the permission to
> spawn a thread. However, I have attempted to change permissions on the
> folder from read only, and I continue to get the same error.
It means that the python script doesn't have permission to execute the
cvs.exe (or possibly the rlog.exe) executable. Please check the
permissions on these executables and their parent directories.
--
- Russell Yanofsky (PGP ID: 0x5FAA0216)
- http://russ.yanofsky.org/
--
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