perl regex - pattern matching

Question

Can anyone explain what is being done below?

 $name=~m,common/([^/]+)/run.*/([^/]+)/([^/]+)$,;

Testing to see if a string matches a regular expression but not doing anything with the result. — Shawn
– Shawn, Commented Nov 2, 2021 at 21:12
The code checks variable $name for a match against common/*/run*/*/*, in term of shell (bash, zsh, csh, ...), perhaps match against to some path in filesystem. It can be rewritten as $name =~ m!common/.*?/run.*?/.*?/.*?$!; — Polar Bear
– Polar Bear, Commented Nov 2, 2021 at 22:37
Re "common/*/run*/*/* in term of shell", I think that's what it was meant to do, but that's not what it does. .* can match strings containing /. /// @PolarBear Re "It can be rewritten as", All those ? are pointless. Your rewrite can match common/foo/run/bar/baz/moo/moo/moo/moo/moo/moo/moo/moo/moo. I don't think you intended that. Remember that ^/.*?/$ matches ///. — ikegami
– ikegami, Commented Nov 2, 2021 at 23:32
@ikegami Indeed I did not intended it, if I would use something of this kind in the code I would perhaps chosen other approach. One line of code does not say much about intentions of the programmer or the problem -- it is not clear what perl script supposed to do. — Polar Bear
– Polar Bear, Commented Nov 3, 2021 at 0:25

ikegami · Accepted Answer · 2021-11-02 23:44:20Z

4

I think it's trying to match a path of one or both of the following forms:

Where

XXX is a sequence of at least one character that doesn't contain /.
YYY is a sequence of any number of characters (incl zero) that doesn't contain /.

It matches more than that, however.

The parts in bold are captured (available to the caller).

answered Nov 2, 2021 at 23:44

ikegami

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