Re: Regular expressions

You're pretty close. I think the problem is that you're being 'greedy' in the regex. I've added a ? after the .* to make it non-greedy. The ?: tells the inner parens to not capture. The outer parens will do this for us.

This example captures everything including the delimiters:

m/(ATG.*?(?:TAG|TAA|TGA))/g
[download]

This one will capture only inside the delimiters:

m/ATG(.*?)(?:TAG|TAA|TGA)/g
[download]

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Re^2: Regular expressions by lairel (Novice) on Oct 27, 2015 at 16:38 UTC
this may be a stupid question, but what is the function of the :	[reply]
Re^3: Regular expressions by stevieb (Canon) on Oct 27, 2015 at 17:02 UTC
As I said, the `?:` makes it so the group within the `()` is not captured. Observe... Without `?:`: `perl -E '"123" =~ /(2\|3)/; say $1' 2` [download] With `?:`: `perl -E '"123" =~ /(?:2\|3)/; say $1'` [download] Note how using the `?:` doesn't put anything into the special numbered `$1` variable. See perlretut's Non-Capturing Groupings	[reply] [d/l] [select]