I want to take a text pattern that occurs the last before another text pattern.
For example I have this text:
code 4ab6-7b5
Another lorem ipsum
Rando
If your regex flavor supports lookaheads, you can use a solution like this
^code:[ ]([0-9a-f-]+)(?:(?!^code:[ ])[\s\S])*id-x
And you can find your result in capture number 1
.
How does it work?
^code:[ ] # match "code: " at the beginning of a line, the square
# brackets are just to aid readability. I recommend always
# using them for literal spaces.
( # capturing group 1, your key
[0-9a-f-]+ # match one or more hex-digits or hyphens
) # end of group 1
(?: # start a non-capturing group; each "instance" of this group
# will match a single arbitrary character that does not start
# a new "code: " (hence this cannot go beyond the current
# block)
(?! # negative lookahead; this does not consume any characters,
# but causes the pattern to fail, if its subpattern could
# match here
^code:[ ] # match the beginning of a new block (i.e. "code: " at the
# beginning of another line
) # end of negative lookahead, if we've reached the beginning
# of a new block, this will cause the non-capturing group to
# fail. otherwise just ignore this.
[\s\S] # match one arbitrary character
)* # end of non-capturing group, repeat 0 or more times
id-x # match "id-x" literally
The (?:(?!stopword)[\s\S])*
pattern let's you match as much as possible without going beyond another occurrence of stopword
.
Note that you might have to use some form of multi-line mode for ^
to match at the beginning of a line. The ^
is important to avoid false negatives, if your random text
contains open:
.
Working demo (using Ruby's regex flavor, as I'm not sure which one you are ultimately going to use)