You don't need to use cat here.

Your greps need some escaping. Right now the . means 'match any character', not 'match .'

And your greps won't return anything because first you reject everything except your start time, and after that you reject everything except your end time. You need to match start and end and everything inbetween. Since grep doesn't understand what date and time means -- or what digits mean, for that matter -- I don't think grep can do what you want all by itself. There's several things with : in it that could match anyway, so that probably won't narrow it down to what you want. I'll work on this a bit...

---------- Post updated at 12:38 PM ---------- Previous update was at 12:25 PM ----------

Here's how you'd match exactly the hours you want.

Code:

#!/bin/sh

# Build a list of the hours we want, to fill into egrep
START=11
END=14
STR=$START
for ((N=START+1; N<=END; N++))
do
        STR="$STR|$N"
done

# This will match 14:06 but not 10:02 since it starts at 11
echo -e "[22/Sep/2009:10:02:24 -0400]\n[22/Sep/2009:14:06:32 -0400]" |
        egrep "/[0-9]+:($STR):"

You could tack another grep to match a specific hostname or what have you. It's not too sophisticated. You can't take it too much farther because grep can't understand what the dates actually mean.

This is about as complex as I'd bother making it in a shell script, since shell in general has a hard time processing date information. If you want something smart enough to just specify a beginning time and date and end time and date, I'd just use perl and process the dates wholesale to compare them.