5

I have below script that I get details of some IPs in a blacklist.

check.sh

blacklist_ips='
    5.56.148.140
    94.73.159.66
    113.171.224.169
    107.150.42.226
    195.159.233.44
    89.19.7.58
'

for ipx in $blacklist_ips
do
    country=`whois $ipx | grep -i country | tail -1 | awk '{print $2}'`
    hostx=`host $ipx |awk '{print $NF}'|sed "s/\.$//"`
    printf '%s %s %s' $country $ipx $hostx          
    printf '\n'
done

It works but I want to sort the output by country section.

Now I can simply go:

bash check.sh | sort -nr

and it works OK.

But I want to do the sorting before sending the result to terminal. In other words sorting should be done inside check.sh and then should be sent to terminal.

How can I achieve this in bash script?

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4
  • 1
    change the print statements to: printf '%s %s %s\n' $country $ipx $hostx > file.txt then add sort -nr file.txt after the for loop? Commented Jul 10, 2016 at 18:30
  • Thanks for the suggestion. I do not want to clutter the folder with another file. Is it possible to do it without file i/o? Commented Jul 10, 2016 at 18:32
  • 1
    How many IP addresses are you looking at? There's only so much a bash script variable can hold. Would simply removing the file at the end of the script work? Commented Jul 10, 2016 at 18:34
  • Well the script can be used by mutliple cron jobs or scripts. So I am a bit afraid of race condition. Do not want to read output of a previous script. Commented Jul 10, 2016 at 18:36

1 Answer 1

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11

Just change the final done to done | sort -nr

This will sort the output of the for loop via a pipe and not need temporary files.

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2
  • Thanks Stephen for your help. How can I add conditional for done | sort -nr please? For some conditionals, I would like to run as done | sort -nrk2? I tried to convert last line of for to a if conditional but it gave syntax error. Commented Jul 10, 2016 at 18:47
  • 3
    Lots of ways. One simple way is to just set a variable order=nr or order=nrk2 before the for loop based on the condition, then you can do sort -$order Commented Jul 10, 2016 at 18:51

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