Shell Programming and Scripting

itry this one also era when i execute it it tell me

Code:

syntax: files...

i think the error might be on the

Code:

case $# in 0) echo syntax: $0 files ...; exit 2;; esac

You invoke it on the files you want to search. If you called the script datefind, you'd run it like

Code:

./datefind traffic/127.0.0.1

and it will prompt you for the date, and search in the file traffic/127.0.0.1 for the given date.

Once again, it doesn't really offer any advantage over regular grep, unless you specifically want the interactive prompting for a date. (Saves you from quoting the date, too.)

Code:

read var
for i in *
do
grep $var $i
if [ $? -le 0 ] 
then
echo $1:`grep $var $i`
fi
done

summer_cherry: you need to double-quote $var, as it might contain spaces.

Also, running grep twice seems kind of excessive. Already the first invocation will print the match (unless you redirect or suppress it), albeit without the "$1": prefix. If you meant to print the file name, that's not entirely correct; you are looping over files with the file name in $i, so that is likely what you intended to print, but you can do that with grep "$var" $i /dev/null or -- in some versions -- grep -H

Anyway, the loop isn't really necessary, because grep can handle multiple input files.

Code:

#!/bin/sh

echo -n 'Type in the date: '
read date
grep "$date" *

This will search all files in the current directory. Perhaps that's what the OP wants, although the problem description mentions a subdirectory with log files, from which (as I understand it) only one is to be searched at a time.