Bash & awk : Segmentation fault while trying to use bash script arguement in awk

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I'm trying to write a script to pull logs between two timestamps using awk.

So this works :

awk '$0 >= "07-Nov-2021 13:40:02.124" && $0 <= "07-Nov-2021 13:43:08.124"' websniffer.log

But when I use it within a bash script, I'm struggling to get it work. Here's my minimal sample script & error :

#!/bin/bash

fromtime=$1
totime=$2
nameoffile="websniffer.log"
awk -v fromtimestamp=$fromtime -v totimestamp=$totime -v filenm=$nameoffile '$0 >= "$fromtimestamp" && $0 <= "$totimestamp"}' $filenm

Now when I execute this script by :

#terminal@root$ ./samplescript.sh "07-Nov-2021 13:40:02.124" "07-Nov-2021 13:43:08.124"

I get this :

 ./samplescript.sh "07-Nov-2021 13:40:02.124" "07-Nov-2021 13:43:08.124"./samplescript.sh "07-Nov-2021 13:40:02.124" "07-Nov-2021 13:43:08.124" ./samplescript.sh: line 6:  2190 Segmentation fault      awk -v fromtimestamp=$fromtime -v totimestamp=$totime -v filenm=$nameoffile '$0 >= "$fromtimestamp" && $0 <= "$totimestamp"}' $filenm

Can someone please help in getting this working ?

awk shell scripting linux bash
1 Answers

There's a couple of things:

#!/bin/bash

fromtime=$1
totime=$2
nameoffile="websniffer.log"
awk -v fromtimestamp="$fromtime" -v totimestamp="$totime" '$0 >= fromtimestamp && $0 <= totimestamp' "$nameoffile"

You need to quote the shell variables you pass to awk (they have whitespace in them) and there's no need to create filenm and pass it to awk since you're not using it within the awk script.

user_13947290
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