hello,

i want to make a script to search the file contents in my home directory by a given date and output me the line that has the date...

read -p "what date ?" vardate
echo $vardate
awk -v d="$vardate" '$0 ~ d{print}' weblog



i find this one that really works its output me the correct line and everything...
but each time i need to go and change the filename in order to find what i want...

I recently responded to a similar question of yours here: Shell script to search for text in a file and copy file

#!/bin/sh

case $# in 0|1) echo "syntax: $0 date files ..." >&2; exit 2;; esac

date=$1
shift

awk -v d="$date" '$0 ~ d' "$@"

thanx for replying era!!!

but i am afraid this does not work.
i think that in the script that u write it checking throught files date..
what i want is to check throught files text content...
i have the traffic of my website... in a directory named traffic...
the traffic directory have files with each webpage ip that visit my website!!
so i want the script to be able to output me the ips that visit my website the date that i ask...

You might think wrong. Try it.

Like I already wrote, this is equivalent to grep date file

For example, grep 2008-04-20 traffic/127.0.0.1 would search for 2008-04-20 in the file traffic/127.0.0.1.

Maybe your logs use a different date format, but you get the idea.

PS. Simpler still awk script, provided your date format doesn't have slashes in it:

#!/bin/sh

case $# in 0|1) echo "syntax: $0 date files ..." >&2; exit 2;; esac

date=$1
shift

awk "/$date/" "$@"

i try it!!!

and it does not work!!
its get me a syntax error..

i used it like this

#!/bin/sh

read -p "what date" vardate
echo $vardate

case $# in 0|1) echo "syntax: $0 date files ..." >&2; exit 2;; esac

date=$1
shift

awk "/$date/" "$@"