如何打印管道第一部分的结果？

Question

I have the following grep:

grep -Po '(?<=PROGRAM\()[^\)]+(?=\))' /home/programs/hello_word.sh 

Wich displays the string between PROGRAM( and ) :

RECTONTER

Then, I need to know if these string extracted is contained in a file, so:

grep -Po '(?<=PROGRAM\()[^\)]+(?=\))' /home/programs/hello_word.sh | xargs -I % grep -e % /home/leherad/pgm_currentdate

File content:

RECTONTER
CORASFE
RENTOASD
UBICARP

If its found, returns the line of /home/leherad/pgm_currentdate, but I want to print the line extracted in the first grep ( RECTONTER ). If not found, then wouldn't return nothing.

There is a simple way to do this, or I should not complicate and would be better build a script and save the first grep in a variable?

Answer 1

You can store it on a variable first:

read -r FIRST < <(exec grep -Po '(?<=PROGRAM\()[^\)]+(?=\))' /home/programs/hello_word.sh) && grep -e "$FIRST" /home/leherad/pgm_currentdate

Update 01

#!/bin/bash
shopt -s nullglob
for FILE in /home/programs/*; do
    read -r FIRST < <(exec grep -Po '(?<=PROGRAM\()[^\)]+(?=\))' "$FILE") && grep -e "$FIRST" /home/leherad/pgm_currentdate && echo "$FIRST"
done

Answer 2

I think a straightforward way to solve this is to use a function.

Also, your grep pattern will match shell comments, which could cause unexpected behavior in your xargs command when there are more than one matches; you might want to take steps to only grab the first match. It's hard to say without actually seeing the input files, so I'm guessing this is either ok or comments are actually the expected place for your target pattern.

Anyway, here's my best guess at a function that would work for you.

get_program() {
  local filename="$1"
  local program="$( grep -m1 -Po '(?<=PROGRAM\()[^\)]+(?=\))' "$filename" )"
  if grep -q -e "$program" /home/leherad/pgm_currentdate; then
    echo $program
    grep -e "$program" /home/leherad/pgm_currentdate
  fi
}

get_program /home/programs/hello_word.sh