Hello everybody, I have a problem with creating a script which allows a single parameter to be passed. Sorry if I'm not expert, I'm new at this and .
The code is
Code:
if [ "$1" = "" ]; then
echo "Something"
exit
fi
...does other things...
grep -i $1 $TEMPUSERS > $USERSFILE
COUNT=`wc -l $USERSFILE`
echo "Found $COUNT users"
There are two errors:
1) the grep $1 is not correctly interpreted because the expanded command doesn't have the ' delimitator.
2) There must be some error on the COUNT assignment too, because it doesn't write out the no of lines in the file.
I tried with ksh and bash shell, but it doesn't seem to be a shell type problem.
I know i'm really at the basics, but some help would be very appreciated
Last edited by Franklin52; 10-22-2009 at 02:50 PM..
Reason: Replaced quote tags with code tags
Is it possible to combine a regular expression with a aritmetical expression? For example, taking a 8-numbers caracter sequece and casting each output of a grep, comparing to a constant.
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Using in code something like: awk '{if ( $2 == "0::" ) print}
source file:
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Discussion started by: vishalgoyal
2 Replies
LEARN ABOUT PLAN9
zgrep
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)
10-22-2009
Regular expression as parameter
