Column without last characters with awk and printf-CodePudding

I have this script:

#!/bin/bash

f_status () {
        systemctl list-units | grep $1 | awk '{ printf("SERVICE STATUS:  %-25s \t %s \t %s \t %s\n",$1,$2,$3,$4) }'
}

f_line() {
        echo "-------------------------------------------------------------------------------------------"
}

echo ""
f_line
f_status "cron"
f_status "ssh"
f_line

This script gives me such a result:

-------------------------------------------------------------------------------------------
SERVICE STATUS:  cron.service                    loaded          active          running
SERVICE STATUS:  ssh.service                     loaded          active          running
-------------------------------------------------------------------------------------------

and I search how to remove ".service" from 3d column.

I tried with substr($i, 0, -8) and ${1:-8}

Does anyone have any idea how to get rid of 8 characters from the end to make it look like this:

-----------------------------------------------------------------------------------
SERVICE STATUS:  cron                    loaded          active          running
SERVICE STATUS:  ssh                     loaded          active          running
-----------------------------------------------------------------------------------

CodePudding user response：

You never need grep when you're using awk since grep 'foo' file | awk '{print 7}' can be written as just awk '/foo/{print 7}' file.

Rather than counting characters, just remove everything starting from the last .:

systemctl list-units |
awk -v tgt="$1" '
    {
        svc = $1
        sub(/\.[^.]*$/,"",svc)
    }
    svc == tgt {
        printf "SERVICE STATUS:  %-25s \t %s \t %s \t %s\n",svc,$2,$3,$4
    }
'

I also tightened up your comparison to avoid false matches if called with a service name that's a subset of some other service name or contains regexp metachars like ..

CodePudding user response：

You need to compute end position based on string length, consider following simple example, let file.txt content be

cron.service
ssh.service

then

awk '{print substr($1,1,length($1)-8)}' file.txt

output

cron
ssh

Explanation: arguments for substr are string, start position, end position, length return number of characters in string.

(tested in gawk 4.2.1)

CodePudding user response：

you can use gsub funtion in awk,like the following:

 systemctl list-units | grep 'cron' | awk '{gsub(".service","",$1); printf("SERVICE STATUS:  %-25s \t %s \t %s \t %s\n",$1, $2 ,$3,$4) }'
SERVICE STATUS:  crond                       loaded      active      running