Odd behavior of String#scan

First off, the problem I am trying to solve can be simplified down
to:

‘abcSTARTdef,ghi,jkl,ENDmno’.scan(/START([^,]*,)*END/)

What I want is [["def,"], ["ghi,"], ["jkl,"]] (or the same thing

without the commas), and I still need a way to achieve this. I can
accomplish it with:

‘abcSTARTdef,ghi,jkl,ENDmno’.scan(/START(.*)END/)[0][0].split(/,/)

But this does two operations where it seems like one should suffice.

Does someone know of a way to do this in a single operation?

Back to the odd behavior, the first expression actually returns

[[“jkl,”]]. I can’t figure out how that is the correct answer by any
reasonable definition of “scan”. However, the equivalent String#match
does the same kind of thing, so I must be missing something. Can
someone please explain this behavior?

Thanks,

- Warren B.

Warren B. wrote:

But this does two operations where it seems like one should suffice.

Does someone know of a way to do this in a single operation?

I would suggest:

‘abcSTARTdef,ghi,jkl,ENDmno’.match(/START(.*)END/)[1].split(’,’)

I don’t know of a way to accomplish it in one step. String#scan attempts
to match the whole regex at multiple places within the string; but you
need the START and END to delimit the substring over which scan
operates.

Back to the odd behavior, the first expression actually returns

[[“jkl,”]]. I can’t figure out how that is the correct answer by any
reasonable definition of “scan”. However, the equivalent String#match
does the same kind of thing, so I must be missing something. Can
someone please explain this behavior?

Because you have this:

([^,],)

The final * allows the group to match multiple times. The MatchData will
hold only the last match however, which is “jkl,”.

Will this do?

s=“abcSTARTdef,ghi,jkl,ENDmno”
s.scan(/START(.*)END/).to_s.split(",")

=> [“def”, “ghi”, “jkl”]