Shell script: insert string below a line depending on line number

user_8417039 0
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I am relatively new to shell scripting so apologies. I have found older questions which insert strings based on a string which appears in a line, but my lines are all of the same format.

I have a file in which I want to insert a string 'insert_text. below each line. I used vim to search for the line break and insert the string below in the following:

%s/\n/\r insert_text.\r/

The above works fine. However, I have 250 lines in my file and 12 specific lines for which I don't wish to insert the string below. I could search for each specific line and remove the string, but I would like to write a shell script that could automate it.

This is as close as I've got

#!/bin/bash
for line in {179, 170, 183, 104, 187, 172, 173, 171, 11, 7, 105, 3}
do
     sed '\n/\r insert_text.\r/'
done

Obviously the above loop makes no sense, and the above code does the opposite of what I want, I want to insert the string into all the other lines.

Is there something similar to -v which would select all other lines in the for loop?

shell bash
3 Answers

You can use this awk command:

awk -v ln='179,170,183,104,187,172,173,171,11,7,105,3' '
   ln !~ "(^|,)" NR "(,|$)"{$0 = $0 "\n insert_text."} 1' file
  • We pass a line numbers where we want to skip adding new text using command line argument ln
  • Using !~ operator we ensure current line number doesn't match this comma delimited string
  • If match fails we print current line and new text
user_5482250

Using sed, b jumps out, a appends.

sed '179b;170b;183b;104b;187b;172b;173b;171b;11b;7b;105b;3b;ainsert_text' file
user_3909130

Using perl:

perl -pe 'BEGIN { %skip = map { $_ + 1 => 1 } 0, 179, 170, 183, 104, 187, 172, 173, 171, 11, 7, 105, 3 } print "\n" unless exists $skip{$.}' foo.txt

(The 0 in the list of lines to not print an extra newline after is important)

user_99521960
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