Delete lines that come after a line with a specific pattern in Shell

This is very trivial with text processing utilities. For example, using sed:

sed '1,/pattern/!d' file

Meaning, match every line from the first one to the one with pattern and delete all the non-matched lines. So, replace pattern with your pattern. If it contains /, you need to escape those characters. For example, if the pattern is pattern-with/character:

sed '1,/pattern-with\/character/!d' file

To actually edit files (rather than print the edited stream to stdout), you can use the -i flag:

sed -i '1,/pattern/!d' file

You can make a backup of the original file by adding an extension for the old file to -i. Take care here - you must not include a space before the extension.

sed -i.backup '1,/pattern/!d' file

sed takes multiple filename arguments. For example, to act on all the non-hidden files in the current directory you could use:

sed -i '1,/pattern/!d' *

Another solution, using awk:

awk '/specific-pattern/{stop=1} stop==0{print}' < input_file >> output_file

While the variable stop is 0 (which is is, by default), awk will print the current line. However, if the current line matches the regular expression /specific-pattern/, then stop will be set to 1. This makes stop==0 untrue, so awk will no longer execute the print statement.

Input is read from input_file and appended to output_file.

If you want to keep the line with the pattern, reverse the two parts of the awk script.

Thank you @Zanna

I found this solution :

for ((i=1;i < $count+1 ;i++)) 

sed -n '/PATTERN/q;p' $i.txt > file_out$i.txt

Thank you