Formatting list of IP addresses into a grep command

Shell Programming and Scripting
1

Hi

I have an input file with a list of random IP addresses, each on a new line. Below is just an example as I omitted the real IP addresses for obvious reasons.

Input: random_ip.txt

10.0.0.1
10.0.0.1
10.0.0.1
10.0.0.1
10.0.0.1
10.0.0.1
10.0.0.1
10.0.0.1
10.0.0.1
10.0.0.1
10.0.0.1
10.0.0.1

Basically what I would like to do is to take all the IP addresses in the input file and pipe them into an output file eg. zgrep_ip.sh which in essence is a formatted zgrep command from the input file.

Output: zgrep_ip.sh

zgrep --no-filename -e 10.0.0.1 -e 10.0.0.1  -e 10.0.0.1  -e 10.0.0.1  -e 10.0.0.1  -e 10.0.0.1  *.gz > zgrep.out

I hope it makes sense :confused:

2 
#! /bin/bash
echo -e 'zgrep --no-filename \c' >> zgrep_ip.sh
while read x; do echo -e "-e $x \c" >> zgrep_ip.sh; done < random_ip.txt
echo -e '*.gz > zgrep.out\c'>> zgrep_ip.sh
3

Thanks for your reply, I get the following error;

: No such file or directoryne 3: random_ip.txt

4

I hope you're not being deliberate here.
What's the name of your input file? The file containing the list of IP's.. Isn't it random_ip.txt ?

5

Correct, the file containing the list of IP's is called random_ip.txt, not sure why your script is not picking it up? I executed your script from within the directory where the random_ip.txt file reside.

6

Hmmm... It works for me. Not sure why it doesn't work for you.
May be, some random control characters have crept in either your script file or input file.

7

Why don't you use

zgrep -F -f random_ip.txt *.gz

Doesn't that work?

8

Thanks, will double check.

---------- Post updated at 12:14 PM ---------- Previous update was at 12:12 PM ----------

Seems to be working fine, thanks for that! :b: