CODEWITHSUNDEEP
regex
FPGA
FPGA

Reputation: 3855

RegEx match but ignore in the result

Consider the following string

'#PI + #C + 1.2'

i am trying to get each word after the hash but also ignore the hash in the result

i use 

#(\S+)

which is fine but i am not sure how to make sure the hash is there but ignore it in the result

Upvotes: 0

Views: 432

Answers (2)

sabhiram
sabhiram

Reputation: 907

I think your answer works, here is how I would go about it:

#([^\s]+)\s*

Which reads like: Match a "#", then capture all non-space chars into the first capture group (there has to be at-least one non-space char). After that, match an arbitrary number of spaces.

The entire reg-ex would not "match" or evaluate to true, unless the "#" is found before a bunch of non-whitespace chars.

Obviously, if you are guaranteed of the form #XXX + #YYY + CCC, you can construct a much more targeted reg-ex to pick the appropriate values into the correct capture groups.

Edit: Just noticed that you said "the # shows up in the result", the "result" of what? Typically the regex match object will return the entire matched string, if you are looking to get the contents of (...) - you will have to access the appropriate capture group of the matched regex. And that is very language dependent. In python you can do:

re_SOMETHING = re.compile("#([^\s]+)\s*", ...)
match = re_SOMETHING.match("#PI + #C + 1.2")
if match:
    pi = match.group(1)
    # pi === "PI"

Upvotes: 1

Farhad Alizadeh Noori
Farhad Alizadeh Noori

Reputation: 2306

I don't know exactly what you mean by ignoring the # in the result. Your regex is doing just that. It matches the # to make sure it's there and then only captures the stuff after it. If you don't want the # to match at all you have to use zero-width assertions or look arounds:

(?<=#)(\S+)

Upvotes: 1

Related Questions