On 6/27/2012 11:25 AM, Tim Daneliuk wrote:
On 06/27/2012 09:25 AM, Aleksandr Miroslav wrote:
hello,
I'm not sure if this is the right forum for this question, but here
goes.
I have the following in a shell script:
#!/bin/sh
if [ "$#" -eq "0" ]; then
find /foo
fi
if [ "$#" -eq "1" ]; then
find /foo | grep -i $1
fi
if [ "$#" -eq "2" ]; then
find /foo | grep -i $1 | grep -i $2
fi
if [ "$#" -eq "3" ]; then
find /foo | grep -i $1 | grep -i $2 | grep -i $3
fi
Is there an easier/shorter way to do this? If there are 15 arguments
supplied on the command line, I don't necessarily want to build 15 if
statements.
Thanks in advance for your answers.
The following solution relies on the fact that you can include multiple
patterns for grep to match with the '-e' argument:
#!/bin/sh
PATTERNS=`echo " $*" | sed s/\ /\ -e\ /g`
find /foo | grep $PATTERNS
Notice that when constructing the $PATTERNS string out of the command
line
args, you have to quote them with a prepended space character. That's
because
the subsequent 'sed' substitution needs to find a space *before* each
argument
which it then replaces with "-e ".
This will build a multi-grep string for any number of arguments (within
reason), functionally a boolean AND search:
#!/bin/sh
final_cmd="find /foo"
while [ $# -gt 0 ]
do
final_cmd="$final_cmd | grep -i $1"
shift
done
$final_cmd
May need quoting changes, but it worked on this sample, with this result:
cmdline: ./testshift "1" 1 2 3 4 5
result: "find /foo | grep -i 1 | grep -i 1 | grep -i 2 | grep -i 3 |
grep -i 4 | grep -i 5"
HTH
Brad
_______________________________________________
[email protected] mailing list
http://lists.freebsd.org/mailman/listinfo/freebsd-questions
To unsubscribe, send any mail to "[email protected]"
