Command to insert lines before first match

Question

I have file with the below info

testing
testing
testing

I want to insert a word(tested) before the first testing word using sed or any linu

Answer 1

Exactly:

sed '0,/testing/s/testing/tested\n&/' file

For lines containing "testing":

sed '0,/.*testing.*/s/.*testing.*/tested\n&/' file

For Lines Starting with "testing"

sed '0,/^testing.*/s/^testing.*/tested\n&/' file

For lines ending with "testing":

sed '0,/.*testing$/s/.*testing$/tested\n&/' file

To update the content of the file with the result add "-i", example:

sed -i '0,/testing/s/testing/tested\n&/' file

Answer 2

This might work for you (GNU sed):

sed -e '/testing/{itested' -e ':a;n;ba}' file

Insert tested before the first match of testing and then use a loop to read/print the remainder of the file.

Or use the GNU specific:

sed '0,/testing/itested' file

Answer 3

To provide an awk-based alternative that is easier to understand:

awk '!found && /testing/ { print "tested"; found=1 } 1' file

• found is used to keep track of whether the first instance of testing has been found (variable found, as any Awk variable, defaults to 0, i.e., false in a Boolean context).
• /testing/ therefore matches the first line containing testing, and processes the associated block:
  • { print "tested"; found=1 } prints the desired text and sets the flag that the first testing line has been found
• 1 is a common shorthand for { print }, i.e., simply printing the current input line as is.