How to insert command output directly into a file before a pattern?

Question

This question and its answers show the content of an existing file ("processedFile") can be inserted into another existing file ("receivingFile") before one of its patterns inside, such that:

sed -i -e '/pattern/r processedFile' receivingFile

At times, such content might result from a previous command-unit, and is not yet written.

At that point, is it possible to insert the outcome of the command-unit into receivingFile without creating/writing processedFile?

Example:

Say, we preprocessed someFile as follows (reading its content starting from the 4th line to the end of file):

awk 'NR>3' someFile

Now, without writing this to a new file, we want to directly insert it into receivingFile before a pattern such that:

awk 'NR>3' someFile | <insertion commands>

PS: Preferably by using awk or sed.

---

Based on the request from @RomanPerekhrest,

someFile:

# random output
# random
11
0 0 0 23.259058 2.1756592 -1.2097659 -0.074128056 0.34343502 -0.017659865
0 0 0.05 51.091269 4.8520157 -0.0011590485 0.00096592555 0.0059719715 -0.050598505

Preprocessing:

awk 'NR>3' someFile :

0 0 0 23.259058 2.1756592 -1.2097659 -0.074128056 0.34343502 -0.017659865
0 0 0.05 51.091269 4.8520157 -0.0011590485 0.00096592555 0.0059719715 -0.050598505

receivingFile :

stackExchange is the best.
pattern
Unix stackExchange is the best of the bests.

Answer 1

I can't tell if it's portable accross all *NIX, depending on implementation. It's working on my Linux Debian.

TL;DR: use /dev/stdin as included file, which is sed's input and will get the output of the command before the pipe, like

awk 'NR>3' someFile | sed -i -e '/pattern/r /dev/stdin' receivingFile

Example: destination file contents called receivingFile:

stackExchange is the best.
pattern
Unix stackExchange is the best of the bests.

Input sample called someFile:

# random output
# random
11
0 0 0 23.259058 2.1756592 -1.2097659 -0.074128056 0.34343502 -0.017659865
0 0 0.05 51.091269 4.8520157 -0.0011590485 0.00096592555 0.0059719715 -0.050598505

To insert after the pattern:

awk 'NR>3' someFile | sed -e '/pattern/r /dev/stdin' receivingFile

To insert before the pattern, reusing the answer linked by OP and dump to output:

awk 'NR>3' someFile | sed -n -e '/pattern/r /dev/stdin' -e 1x -e '2,${x;p}' -e '${x;p}' receivingFile

To do what was asked and also replace the file:

awk 'NR>3' someFile | sed -i -n -e '/pattern/r /dev/stdin' -e 1x -e '2,${x;p}' -e '${x;p}' receivingFile

New content of receivingFile:

stackExchange is the best.
0 0 0 23.259058 2.1756592 -1.2097659 -0.074128056 0.34343502 -0.017659865
0 0 0.05 51.091269 4.8520157 -0.0011590485 0.00096592555 0.0059719715 -0.050598505
pattern
Unix stackExchange is the best of the bests.

UPDATE: as noted by @don_crissti, this doesn't work if pattern is the last line. Here's an other command, adapted from his answer to the same linked question! Some other command is needed to accomodate with the temporary addition (and suppression by sed) of the extra blank line. These two would work:

awk 'NR>3' someFile | sed -e '/pattern/{r/dev/stdin' -e 'N;:l;$!n;$!bl};${/^$/!{s/\n$//};//d}' receivingFile <(printf '\n') > receivingFile.new
mv receivingFile.new receivingFile

or just (not needing bash and leaving sed -i to handle the final file):

awk 'NR>3' someFile | { echo '' >> receivingFile; sed -i -e '/pattern/{r/dev/stdin' -e 'N;:l;$!n;$!bl};${/^$/!{s/\n$//};//d}' receivingFile ; }

Look at the linked answer for the line-splitted sed code.

Answer 2

Just use ex, the POSIX-specified scriptable form of vi (which is also POSIX-specified).

printf '%s\n' '/pattern/-r !awk "NR>3" somefile' x | ex receivingFile

Or more generally:

printf '%s\n' '/pattern/-r !somecommand' x | ex somefile

Answer 3

With sed (GNU sed) 4.4

sed -i '/pattern/{h;s/.*/awk "NR>3" somefile/e;p;x}' receivingFile