awk print lines before while match INFO untill match ERROR - regex

Question

I want to print lines before my /ERROR/ match. The lines to be printed should be all containing INFO untill the previous ERROR is found.
So If I had a file
ERROR this is an error
INFO error found on line 2
INFO error is due to something
ERROR this is another error
I want the /ERROR/ from ERROR this is another error to print
INFO error found on line 2
INFO error is due to something
ERROR this is another error
Anyone know?
Part of my current script:
/CRITICAL/ {
print "\x1b[93;1m"
}
/ERROR/ {
print "\x1b[37m"
}
/ERROR|EMERGENCY|CRITICAL/ {
if (NR == n+1) print "";
n = NR;
print x;print
print "\x1b[0m"
};{x=$0}'

Answer 1

Try this one liner:
awk 'x;/ERROR/{x=1}' file
Out:
INFO error found on line 2
INFO error is due to something
ERROR this is another error
Long version:
x;/ERROR/{
x1=1;
print
}
If "ERROR" is found x=1, if x is true we've already gone through that line, then we print until we pass that line again.
Or maybe this, I don't have very clear what output you need.
awk '/ERROR/{x=1;next}/ERROR/{x=1}x'
Out
INFO error found on line 2
INFO error is due to something