Shell Programming and Scripting

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: https://www.unix.com/unix-dummies-que...#post302187223

Code:

#!/bin/sh

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

date=$1
shift

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

This expects the date as the first parameter, and a list of files as the remaining parameters. Those are passed through to awk in "$@" after the first argument (the date) has been shifted off and passed to the awk script as a variable.

I took out the { print } because that is the default action; this is probably less readable, so if you don't use awk much, it might be safer to leave it in.

Of course, this simple script is exactly equivalent to grep without any options or other bells & whistles.

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:

Code:

#!/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

Code:

#!/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 "/$date/" "$@"

am i wrong somewhere??


my log files text are in this format : 162.12.56.7 Tues Feb 8 21:02:35 GMT 2008

What's the syntax error? And why did you add the "read vardate" stuff? You're supposed to pass in the date on the command line. Of course, you can change that, but then you need to modify the rest of the script accordingly.

When passing a value with spaces in it, you need to use quotes:

Code:

grep "Tues Feb 8" traffic/127.0.0.1

Here's another try at a script:

Code:

#!/bin/sh

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

echo enter date:
read date

grep "$date" "$@"

The script will be more useful for use from other scripts if you let it accept the date on the command line rather than read it interactively, but that may not be an issue here. Anyway, you need to pass it the file(s) to search as arguments anyway, so why not make the date a command-line argument, too (or else ask which files to grep, interactively).