is file open in system ? - other than lsof

[Nick Craig-Wood]

Thomas Guettler <hv at tbz-pariv.de> wrote:
>  bvidinli schrieb:
> > is there a way to find out if file open in system ? -
> > please write if you know a way  other than lsof. because lsof if slow for me.
> > i need a faster way.
> > i deal with thousands of files... so, i need a faster / python way for this.
> > thanks.
> 
>  On Linux there are symlinks from /proc/PID/fd to the open
>  files. You could use this:
> 
>  #!/usr/bin/env python
>  import os
>  pids=os.listdir('/proc')
>  for pid in sorted(pids):
>       try:
>           int(pid)
>       except ValueError:
>           continue
>       fd_dir=os.path.join('/proc', pid, 'fd')
>       for file in os.listdir(fd_dir):
>           try:
>               link=os.readlink(os.path.join(fd_dir, file))
>           except OSError:
>               continue
>           print pid, link

Unfortunately I think that is pretty much exactly what lsof does so is
unlikely to be any faster!

However if you have 1000s of files to check you only need to do the
above scan once and build a dict with all open files in whereas lsof
will do it once per call so that may be a useful speedup.

Eg

------------------------------------------------------------
import os
pids=os.listdir('/proc')
open_files = {}
for pid in sorted(pids):
     try:
         int(pid)
     except ValueError:
         continue
     fd_dir=os.path.join('/proc', pid, 'fd')
     try:
         fds = os.listdir(fd_dir)
     except OSError:
         continue
     for file in fds:
         try:
             link=os.readlink(os.path.join(fd_dir, file))
         except OSError:
             continue
         if not os.path.exists(link):
             continue
         open_files.setdefault(link, []).append(pid)

for link in sorted(open_files.keys()):
    print "%s : %s" % (link, ", ".join(map(str, open_files[link])))
------------------------------------------------------------


You might want to consider http://pyinotify.sourceforge.net/ depending
on exactly what you are doing...

-- 
Nick Craig-Wood <nick at craig-wood.com> -- http://www.craig-wood.com/nick