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Operating Systems Solaris A strange user appears in my quotas and I can't find it in my system Post 302280771 by jlliagre on Tuesday 27th of January 2009 02:13:13 PM
Old 01-27-2009
Quote:
Originally Posted by lzcool
I've used the command find / -name "833" but it doesn't return any matches.
That should be:
Code:
find / -user 883

 

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LESSECHO(1)						      General Commands Manual						       LESSECHO(1)

NAME
lessecho - expand metacharacters SYNOPSIS
lessecho [-ox] [-cx] [-pn] [-dn] [-mx] [-nn] [-ex] [-a] file ... DESCRIPTION
lessecho is a program that simply echos its arguments on standard output. But any metacharacter in the output is preceded by an "escape" character, which by default is a backslash. OPTIONS
A summary of options is included below. -ex Specifies "x", rather than backslash, to be the escape char for metachars. If x is "-", no escape char is used and arguments con- taining metachars are surrounded by quotes instead. -ox Specifies "x", rather than double-quote, to be the open quote character, which is used if the -e- option is specified. -cx Specifies "x" to be the close quote character. -pn Specifies "n" to be the open quote character, as an integer. -dn Specifies "n" to be the close quote character, as an integer. -mx Specifies "x" to be a metachar. By default, no characters are considered metachars. -nn Specifies "n" to be a metachar, as an integer. -fn Specifies "n" to be the escape char for metachars, as an integer. -a Specifies that all arguments are to be quoted. The default is that only arguments containing metacharacters are quoted SEE ALSO
less(1) AUTHOR
This manual page was written by Thomas Schoepf <schoepf@debian.org>, for the Debian GNU/Linux system (but may be used by others). Send bug reports or comments to bug-less@gnu.org. Version 487: 25 Oct 2016 LESSECHO(1)
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