Search for value and print something if found (BASH) - bash

Question

I have the following list:
COX1
COX1
COX1
COX1
COX1
Cu-oxidase
Cu-oxidase_3
Cu-oxidase_3
Fer4_NifH
and I want to search if COX1 and Cu-oxidase is in the list, I want to print xyz, if Cu-oxidase_3 and Fer4_NifHis in the list too (independent if the first two are in the list, then it should print abc.
This is what I could script so far:
if grep 'COX1' file.txt; then echo xyz; else exit 0; fi
but it is of course incomplete.
Any solution to that?
ideally my output would be:
xyz
abc

Answer 1

Awk lets you easily search for multiple regular expressions and print something else than the matched string itself. (grep can easily search for multiple patterns, too, but it will print the match or its line number or file name, not some arbitrary string.)
The following assumes that you have a single token per line. This assumption makes the script really simple, though it would also not be hard to support other scenarios.
awk '{ a[$1]++ }
END { if (("COX1" in a) && ("Cu-oxidase" in a)) print "xyz";
if (("Cu-oxidase_3" in a) && ("Fer4_NifH" in a)) print "abc" }' file.txt
This builds an associative array of each token (actually the first whitespace-separated token on each line) and then at the end, when it has read every line in the file, checks whether the sought tokens exist as keys in the array.
Performing a single pass over the input file is a big win especially if you have a large input file and many patterns. Just for completeness, the syntax for performing multiple passes with grep is very straightforward;
if grep -qx 'COX1' file.txt && grep -qx 'Cu-oxidase' file.txt
then
echo xyz
fi
which can be further abbreviated to
grep -qx 'COX1' file.txt && grep -qx 'Cu-oxidase' file.txt && echo xyz
Notice the -x switch to require the whole line to match (otherwise the regex 'Cu-oxidase' would also match on the Cu-oxidase_3 lines).

Answer 2

Above is a very verbose way to achieve this. There are ways to write the same with less ifs and less greps, but I really wanted to show you the logic:
you run a grep command, check for its return value with $?, and finally acts on the conditions.
# default values
HAS_COX1=0
HAS_CUOX=0
HAS_CUO3=0
HAS_FER4=0
# run silently grep
grep -q 'COX1' file.txt
# check for return value and set variable accordingly
if [ $? -eq 0 ]; then HAS_COX1=1; fi
# same as above
grep -q 'Cu-oxidase' file.txt
if [ $? -eq 0 ]; then HAS_CUOX=1; fi
grep -q 'Cu-oxidase_3' file.txt
if [ $? -eq 0 ]; then HAS_CUO3=1; fi
grep -q 'Fer4_NifH' file.txt
if [ $? -eq 0 ]; then HAS_FER4=1; fi
if [ $HAS_COX1 -eq 1 ]; then
if [ $HAS_CUOX -eq 1 ]; then
echo 'xyz'
exit 0
fi
fi
if [ $HAS_CUO3 -eq 1 ]; then
if [ $HAS_FER4 -eq 1 ]; then
echo 'abc'
exit 0
fi
fi
echo 'None of the checks where matched'
exit 1
Beware: this code is untested, so there might be bugs ☺
The code isn't perfect, as it cannot print both 'xyz' and 'abc' when both conditions are met (but that would be an easy fix with the syntax I provide). Also $HAS_CUOX will be set to 1 whenever $HAS_CUO3 is found (no boundary checking in the grep regex).
You could take that code further by using a single grep for each set of conditions to check, using something like 'COX1\|Cu_oxidase' as the regex for grep. And also fix the minor issues I mentioned above.
ideally my output would be:
xyz
abc
You added your expected output after I wrote the above script, but given the elements I gave you, you should be able to figure how to improve that (basically removing the exit 0 where I placed them, and doing exit 1 when no output has been given.
Or just remove all exits as a dirty solution.