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hiding output from find command

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# 1  
Old 01-12-2008
hiding output from find command
when I do the find command from / , there are a lot of directories that I do not have access to and so I get

"find: cannot open ..."

How can I suppress these messages so only what was found is output.

I was thinking on

Code:
find / -name 'searchterm' | grep -v find

but this doesnt work

Thanks
# 2  
Old 01-12-2008
If you use SH, then you can direct stderr, which is what is causing your "cannot open.." messages, to /dev/null and suppress it that way ie:
Code:
find / -name 'searchterm'   2> /dev/null

# 3  
Old 01-13-2008
Quote:
Originally Posted by redhead
If you use SH, then you can direct stderr, which is what is causing your "cannot open.." messages, to /dev/null and suppress it that way ie:
Code:
find / -name 'searchterm'   2> /dev/null

that is one way to do it. Thanks.

But just out of curiosity, why does the grep -v not find not work?

if we do

Code:
ls -ltr | grep -v find

then all the files containing find will not be shown.

If we do

Code:
find / -name 'filename' | grep -v find

then I would expect all lines returned with "find" on them being filtered out.

Why does this not happen?

Thanks.
# 4  
Old 01-13-2008
I believe its because the "find: cannot open ..."

is going to STDERR and

and you are piping the STDOUT to grep -v
rubio
# 5  
Old 01-13-2008
Quote:
Originally Posted by JamesByars
Code:
find / -name 'filename' | grep -v find

then I would expect all lines returned with "find" on them being filtered out.

Why does this not happen?

Thanks.
That's because only stdout is being piped to grep while stderr is being sent to the terminal. In order to exclude stuff using grep send stderr to the same place as stdout.

Code:
find / -name 'filename' 2>&1 | grep -v find

# 6  
Old 01-13-2008
that makes complete sense, thanks
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