bash：在标准输出上触发输出

Question

I run a program which writes a lot to stdout . After a certain time this program prints a defined line to stdout and I need to trigger this in order to call a function parallel (without terminating the first program).

How can I do this in bash?

---

A bit more explained: I need to run an installation program which is an executable from a mounted dvd1.iso . After some time it prints "Info: Eject DVD 1 and insert DVD 2 to continue.". And this is what shall be done automatically.

---

---

Following the answer here my test set up:

talker.sh

#!/bin/bash

for VAR in {1..20}
do
    sleep 1s
    echo "huhu $VAR"
done

listener.sh

#!/bin/bash

bash talker.sh \
    | tee output.txt \
    | grep --line-buffered "huhu 3" \
    | ( while read -r line; do echo "found"; done; ) &\
    tail -f output.txt

and how it works:

$ bash listerner.sh 
huhu 1
huhu 2
huhu 3
found
huhu 4
huhu 5
...

Answer 1

You could save the output of the first program in a file with tee and, at the same time, filter the output to get the desired pattern. Something like:

./program | tee output.txt | grep --line-buffered pattern | ( while read -r line; do something; done; )

As suggested in a comment below by @thatotherguy the option --line-buffered should prevent grep to hold on to the matches.