Insert text before a certain line using Bash

Question

How can I insert a set of lines (about 5) into a file at the first place a string is found?

For example:

BestAnimals.txt

dog         


        

Answer 1

An awk solution:

awk '/cat/ && c == 0 {c = 1; print "giraffe"}; {print}' \
     BestAnimals.txt

If the animals you want to insert are in "MyOtherBestAnimals.txt" you can also do

awk '/cat/ && c == 0 {c = 1; system("cat MyOtherBestAnimals.txt") }; {print} ' \
     BestAnimals.txt

This answer can basically be broken down as follows, because ; separates the awk condition-action pairs:

• /cat/ && c == 0 { c = 1; ... } sets c to 1 at the first row containing cat. The commands put at the ... are then executed, but only once, because c is 1 now.
• {print} is the action print with no condition: prints any input line. This is done after the above condition-action pair.

Depending on what is actually at the ..., giraffe is printed, or the contents of "MyOtherBestAnimals.txt" is sent to the standard output, before printing the first line containing "cat".

Edit

After analysis of @glenn jackman's solution, it seems this solution can still be improved: when using input file

nyan cat
cat

the data is appended before nyan cat and not before the line equal to cat. The solution is then to request the full line to be equal to cat:

awk '$0 == "cat" && c == 0 {c = 1; print "giraffe"}; {print}' \
     BestAnimals.txt

for the insertion of a single line and

awk '$0 == "cat" && c == 0 {c = 1; system("cat MyOtherBestAnimals.txt") }; {print} ' \
     BestAnimals.txt

for the insertion of a file