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Top Forums Shell Programming and Scripting hiding output from find command Post 302157911 by shamrock on Sunday 13th of January 2008 03:32:46 PM
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

 

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ZGREP(1)						      General Commands Manual							  ZGREP(1)

NAME
zgrep - search possibly compressed files for a regular expression SYNOPSIS
zgrep [ grep_options ] [ -e ] pattern filename... DESCRIPTION
Zgrep invokes grep on compressed or gzipped files. These grep options will cause zgrep to terminate with an error code: (-[drRzZ]|--di*|--exc*|--inc*|--rec*|--nu*). All other options specified are passed directly to grep. If no file is specified, then the standard input is decompressed if necessary and fed to grep. Otherwise the given files are uncompressed if necessary and fed to grep. If the GREP environment variable is set, zgrep uses it as the grep program to be invoked. EXIT CODE
2 - An option that is not supported was specified. AUTHOR
Charles Levert (charles@comm.polymtl.ca) SEE ALSO
grep(1), gzexe(1), gzip(1), zdiff(1), zforce(1), zmore(1), znew(1) ZGREP(1)
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