regsubst and regex

[Antidot SAS]

Hi everyone,

A quick question for you, here is the code:

$tt=regsubst("test","^(.*)->(.*)",'\2')

Doesn't return: undef or nil, it does return:  "test"

Niether does $tt=regsubst("test","^(.?)->(.*)",'\2') or $tt=regsubst("test","^(.+?)->(.*)",'\2').

Is there a way to return undef if the string doesn't include '->something' ?

Regards,

JM

[Thomas Bellman]

On 2012-04-16 15:26, Antidot SAS wrote:

> A quick question for you, here is the code:
> $tt=regsubst("test","^(.*)->(.*)",'\2')
>
> Doesn't return: undef or nil, it does return: "test"

As expected and intended.

> Niether does $tt=regsubst("test","^(.?)->(.*)",'\2')
> or $tt=regsubst("test","^(.+?)->(.*)",'\2').
>
> Is there a way to return undef if the string doesn't include '->something' ?

Perhaps something like this:

$x = 'test'
$temp = regsubst($x, '^(.*)->(.*)', '\2')
if $temp == $x {
# No substitution done, must mean there is no "->" in $x
$tt = undef
} else {
$tt = $temp
}

Or perhaps this:

$tt = regsubst($x, '^(.*)->(.*)|.*', '\2')

It will however give you the empty string (""), not undef, if there is
no "->" in the string.

/Bellman

[Antidot SAS]

Thx for the reply it helps.

But how come the \2 returns something that I never asked?

       /Bellman

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[Thomas Bellman]

On 2012-04-16 15:57, Antidot SAS top-posted:

> Thx for the reply it helps.
>
> But how come the \2 returns something that I never asked?

It doesn't. But since there are no occurrances of '^(.*)->(.*)',
then there are none that get replaced. Similarly, if you do:

regsubst('foobar', 'x', 'y')

you will get 'foobar', since there *is* no 'x' in 'foobar' that
can be replaced by a 'y'.

/Bellman

[Antidot SAS]

Ok, I totally forget that replacement part is not a search :D

       /Bellman