Re: Howto strip 42 from comma separated list using s///

I wouldn't do it with a single substitution, but either with two:

    s/\b42\b,?// && s/,$//;
[download]

or none:

    join ",", grep {$_ != 42} split /,/, $string
[download]

Perl --((8:>*

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Re^2: Howto strip 42 from comma separated list using s/// by ady (Deacon) on Nov 03, 2005 at 13:43 UTC
I'd much prefer one plain vanilla regex (w/o Perl /e magic), so does such a beast exist...? /allan	[reply]
Re^3: Howto strip 42 from comma separated list using s/// by Perl Mouse (Chaplain) on Nov 03, 2005 at 14:05 UTC
Just by listing all cases. `s/,42\b\|\b42,\|^42$//` [download] But next time, ask VB questions on a VB forum. `Perl --((8:>*`	[reply] [d/l]