Possible Duplicate:
check what files are open in Python
Hello,
Is it possible to obtain a list of all currently open file handles, I presume that they are stored somewher开发者_运维百科e in the environment.
I am interested in theis function as I would like to safely handle any files that are open when a fatal error is raised, i.e. close file handles and replace potentially corrupted files with the original files.
I have the handling working but without knowing what file handles are open, I am unable to implement this idea.
As an aside, when a file handle is initialised, can this be inherited by another imported method?
Thank you
lsof, /proc/pid/fd/
The nice way of doing this would be to modify your code to keep track of when it opens a file:
def log_open( *args, **kwargs ):
print( "Opening a file..." )
print( *args, **kwargs )
return open( *args, **kwargs )
Then, use log_open
instead of open
to open files. You could even do something more hacky, like modifying the File
class to log itself. That's covered in the linked question above.
There's probably a disgusting, filthy hack involving the garbage collector or looking in __dict__
or something, but you don't want to do that unless you absolutely really truly seriously must.
If you're using python 2.5+ you can use the with
keyword (though 2.5 needs `from future import with_statement)
with open('filename.txt', 'r') as f:
#do stuff here
pass
#here f has been closed and disposed properly - even with raised exceptions
I don't know what kind of catastrophic failure needs to bork the with
statement, but I assume it's a really bad one. On WinXP, my quick unscientific test:
import time
with open('test.txt', 'w') as f:
f.write('testing\n')
while True:
time.sleep(1)
and then killing the process with Windows Task Manager still wrote the data to file.
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