sed with variable as argument in bash script

Question

I am trying to write a bash script to scan for authorized_keys files and remove the keys of a couple previous employees if found. I am having one heck of a time figuring out the escaping for the sed command at the end. I am using commas instead of / since / can show up in the ssh-key. Any help would be appreciated

#!/bin/bash

declare -A keys
keys["employee1"]='AAAAB3NzaC1yc2EAAAABJQAAAIEAxoZ7ZdpJkL98n8cSTkFBwaAeSNK0m/tOWtF1mu5NAzMM/+1SDO6rJH/ruyyqBJo9s+AHWZLGRHfXT2XBg2SRaUnubAKp0w6qNIbej0MsA/ifAs8AfVGdj0pUPLtKpo6XVZkB8vEZSIQ+xNk1n5HJrGJnFGWKWeY3z1/KOLxcLHU='
keys["employee2"]='AAAAB3NzaC1yc2EAAAABIwAAAQEAwHYNAVhb319OBVXPhYF8cSTkFBwaAekr7UcKjfLPCHMpz19W0L/C0g+75Hn8COxOQILDUhIPhYHXOduQjGD/6NXgJDWxgyT00Azg5BREUnBd58WqZPlEvTZYlAgmdMIbnWPPGdJwzqKH/k7/STK6vTKxL6rxBo4lSNK0m/tOWtF1mu5NAzMM/+1SDO6rJH/ruyyqBJo9s+NIbej0MsA/ifAs8AfAkfO2JjgeQpJMyZ7B02XVN5iSLAyC3Cb5FXIjJuk4LPhcApuVyszH2lgve0r5bt/nFgVujJTvJTHPlGrqkYDcDJVUtfbjoLqGPrnpijp6rGIC7aFDDe7bk0ygHYMXDFWcjJBerfLGUWTYWFFLY3bfiO/h/9oEycmQHyB2co4a0IyyDnaYn9OY6xsRRATVlk4Q=='

files=`find / -name authorized_keys`

echo "Checking Authorized_Keys files on: " `hostname`
echo ""
echo "Located files: "

for file in $files; do
  echo "  $file"
done

echo""

for file in $files; do
  for key in "${!keys[@]}"; do
    if grep -q ${keys[$key]} $file; then
      echo "  *** Removing $key from $file"
      sed "s,${keys[$key]},d" $file
    fi
  done
done

anubhava · Accepted Answer · 2015-06-03 05:52:39Z

2

You've made it a bit complicated I think.

You can do this using grep -vf and process substitution:

# array to hold the value you want to remove
keys=(
'AAAAB3NzaC1yc2EAAAABJQAAAIEAxoZ7ZdpJkL98n8cSTkFBwaAeSNK0m/tOWtF1mu5NAzMM/+1SDO6rJH/ruyyqBJo9s+AHWZLGRHfXT2XBg2SRaUnubAKp0w6qNIbej0MsA/ifAs8AfVGdj0pUPLtKpo6XVZkB8vEZSIQ+xNk1n5HJrGJnFGWKWeY3z1/KOLxcLHU=',
'AAAAB3NzaC1yc2EAAAABIwAAAQEAwHYNAVhb319OBVXPhYF8cSTkFBwaAekr7UcKjfLPCHMpz19W0L/C0g+75Hn8COxOQILDUhIPhYHXOduQjGD/6NXgJDWxgyT00Azg5BREUnBd58WqZPlEvTZYlAgmdMIbnWPPGdJwzqKH/k7/STK6vTKxL6rxBo4lSNK0m/tOWtF1mu5NAzMM/+1SDO6rJH/ruyyqBJo9s+NIbej0MsA/ifAs8AfAkfO2JjgeQpJMyZ7B02XVN5iSLAyC3Cb5FXIjJuk4LPhcApuVyszH2lgve0r5bt/nFgVujJTvJTHPlGrqkYDcDJVUtfbjoLqGPrnpijp6rGIC7aFDDe7bk0ygHYMXDFWcjJBerfLGUWTYWFFLY3bfiO/h/9oEycmQHyB2co4a0IyyDnaYn9OY6xsRRATVlk4Q=='
)

while IFS= read -d '' -r file; do
    grep -vf <(printf "%s\n" "${keys[@]}") "$file" > "$file.tmp"
    mv "$file.tmp" "$file"
done < <(find / -name authorized_keys -print0)

edited Jun 3, 2015 at 5:52

answered Jun 3, 2015 at 4:49

anubhava

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3 Comments

digitaladdictions Over a year ago

Im getting the following error. ./remove.sh: line 9: syntax error near unexpected token <(find / -name authorized_keys -print0)' ./remove.sh: line 9: done <(find / -name authorized_keys -print0)'

marbu Over a year ago

Just a note: directly piping output of find command into while loop would work as well.

anubhava Over a year ago

Yes that will work but then it will fork an unnecessary subshell.

WKPlus · Accepted Answer · 2015-06-03 06:31:05Z

2

In your case, it's easy, just need to use a sign which not contained in base64 code as the delimiter, eg |:

sed "\|${keys[$key]}|d" $file

Explanation in the sed manual:

\%regexp%

(The % may be replaced by any other single character.)

This also matches the regular expression regexp, but allows one to use a different delimiter than /.

edited Jun 3, 2015 at 6:31

answered Jun 3, 2015 at 6:23

WKPlus

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