Create pattern list dynamically from file for running in grep -f

Question

Lets say I have this input

11
12
31
40
42
90

I need to get all number lines from the file and add the | character after each that can be used as a pattern list to run with grep -f .

I was thinking of using xargs but I have no idea how to use | as a separator or put the args in the middle of a string

grep -P "(11|12|31|40|42|90)\|" o.txt

Answer 1

You can use

grep -E -f <(sed -E '/^[0-9]+$/!d;s/.*/&\\|/' numbers.txt) file.txt

First, get all lines that only contain digit from the numbers.txt , append to each found line a \| , and then use this as a pattern file with -f option to grep . The file.txt is a file where you are looking for matches.

See an online demo :

#!/bin/bash
text='Some 11 number and 11| that we need
Redundant line'
s="11
12
31
40
42
90"
grep -E -f <(sed -E '/^[0-9]+$/!d;s/.*/&\\|/' <<< "$s") <<< "$text"
# => Some 11 number and 11| that we need
grep -oE -f <(sed -E '/^[0-9]+$/!d;s/.*/&\\|/' <<< "$s") <<< "$text"
# => 11|