Regex: get the first match

Trochalakis_Christos · 10 June 2007 10:50

Hello!

I want to parse a tagged string like this: "this ismy
string"

i am doing:

"this ismy string".scan(/(.*)<\/i>/)

=> [["this ismy string"]]

What i want is a regex that will return the *first* segment that
matches.
in the above case -> [["this is", "my string"]]

Is there any way to do this?

Thanks!

Forum · 10 June 2007 12:08

This is a FAQ, and yes I will give the solution
Regexps are gready par default, they consume as many chars as
possible, there are some possibilities - not tested:

(1) use non gready matches
"this ismy string".scan(/(.*?)<\/i>/)
(2) use less general expressions
"this ismy string".scan(/(.[^<]*)<\/i>/)
(3) Combine both
"this ismy string".scan(/(.[^<]*?)<\/i>/)

HTH
Robert

P.S.
This *really* is a FAQ though

···

On 6/10/07, Trochalakis Christos <yatiohi@ideopolis.gr> wrote:

Hello!

I want to parse a tagged string like this: "this ismy
string"

i am doing:

>> "this ismy string".scan(/(.*)<\/i>/)
=> [["this ismy string"]]

What i want is a regex that will return the *first* segment that
matches.
in the above case -> [["this is", "my string"]]

Is there any way to do this?

Thanks!

--
You see things; and you say Why?
But I dream things that never were; and I say Why not?
-- George Bernard Shaw

GrzechG · 10 June 2007 12:25

I want to parse a tagged string like this: "this ismy
string"

i am doing:

"this ismy string".scan(/(.*)<\/i>/)

=> [["this ismy string"]]

What i want is a regex that will return the *first* segment that
matches.
in the above case -> [["this is", "my string"]]

The solution is :

"this ismy string".scan(/(.*?)<\/i>/)
=> [["this is"], ["my string"]]

The regexp scope is default maximum as is possible to find.
If you use '?' character you minimze the scope.
(.*?) instead of (.*) and the part of string don't be include into one result.

Regards,
Grzegorz Golebiowski

Logan_Capaldo · 10 June 2007 12:21

> Hello!
>
> I want to parse a tagged string like this: "this ismy
> string"
>
> i am doing:
>
> >> "this ismy string".scan(/(.*)<\/i>/)
> => [["this ismy string"]]
>
> What i want is a regex that will return the *first* segment that
> matches.
> in the above case -> [["this is", "my string"]]
>
> Is there any way to do this?
>
> Thanks!
>
This is a FAQ, and yes I will give the solution
Regexps are gready par default, they consume as many chars as
possible, there are some possibilities - not tested:

(1) use non gready matches
"this ismy string".scan(/(.*?)<\/i>/)
(2) use less general expressions
"this ismy string".scan(/(.[^<]*)<\/i>/)
(3) Combine both
"this ismy string".scan(/(.[^<]*?)<\/i>/)

.Unless you want to match strings like <foo, it would be simple to
just use [^<]*, and not .[^<]*. .[^<]* will also not match . If the
intent was to make the regexp not match that, a better regexp would be [^<]+

HTH

···

On 6/10/07, Robert Dober <robert.dober@gmail.com> wrote:

On 6/10/07, Trochalakis Christos <yatiohi@ideopolis.gr> wrote:
Robert

P.S.
This *really* is a FAQ though
--
You see things; and you say Why?
But I dream things that never were; and I say Why not?
-- George Bernard Shaw

Trochalakis_Christos · 10 June 2007 17:20

Thanks Grzegorz, nice trick!

···

On Jun 10, 3:22 pm, GrzechG <grze...@DELITgazeta.pl> wrote:

> I want to parse a tagged string like this: "this ismy
> string"

> i am doing:

>>> "this ismy string".scan(/(.*)<\/i>/)
> => [["this ismy string"]]

> What i want is a regex that will return the *first* segment that
> matches.
> in the above case -> [["this is", "my string"]]

The solution is :

"this ismy string".scan(/(.*?)<\/i>/)
=> [["this is"], ["my string"]]

The regexp scope is default maximum as is possible to find.
If you use '?' character you minimze the scope.
(.*?) instead of (.*) and the part of string don't be include
into one result.

Regards,
Grzegorz Golebiowski

Forum · 10 June 2007 12:45

Thanks for correcting my typos.
Robert

···

On 6/10/07, Logan Capaldo <logancapaldo@gmail.com> wrote:

On 6/10/07, Robert Dober <robert.dober@gmail.com> wrote:
>
> On 6/10/07, Trochalakis Christos <yatiohi@ideopolis.gr> wrote:
> > Hello!
> >
> > I want to parse a tagged string like this: "this ismy
> > string"
> >
> > i am doing:
> >
> > >> "this ismy string".scan(/(.*)<\/i>/)
> > => [["this ismy string"]]
> >
> > What i want is a regex that will return the *first* segment that
> > matches.
> > in the above case -> [["this is", "my string"]]
> >
> > Is there any way to do this?
> >
> > Thanks!
> >
> This is a FAQ, and yes I will give the solution
> Regexps are gready par default, they consume as many chars as
> possible, there are some possibilities - not tested:
>
> (1) use non gready matches
> "this ismy string".scan(/(.*?)<\/i>/)
> (2) use less general expressions
> "this ismy string".scan(/(.[^<]*)<\/i>/)
> (3) Combine both
> "this ismy string".scan(/(.[^<]*?)<\/i>/)

.Unless you want to match strings like <foo, it would be simple to
just use [^<]*, and not .[^<]*. .[^<]* will also not match . If the
intent was to make the regexp not match that, a better regexp would be [^<]+

--
You see things; and you say Why?
But I dream things that never were; and I say Why not?
-- George Bernard Shaw

Forum · 10 June 2007 17:31

You are welcome
Robert

···

On 6/10/07, Trochalakis Christos <yatiohi@ideopolis.gr> wrote:

On Jun 10, 3:22 pm, GrzechG <grze...@DELITgazeta.pl> wrote:
> > I want to parse a tagged string like this: "this ismy
> > string"
>
> > i am doing:
>
> >>> "this ismy string".scan(/(.*)<\/i>/)
> > => [["this ismy string"]]
>
> > What i want is a regex that will return the *first* segment that
> > matches.
> > in the above case -> [["this is", "my string"]]
>
> The solution is :
>
> "this ismy string".scan(/(.*?)<\/i>/)
> => [["this is"], ["my string"]]
>
> The regexp scope is default maximum as is possible to find.
> If you use '?' character you minimze the scope.
> (.*?) instead of (.*) and the part of string don't be include
> into one result.
>
> Regards,
> Grzegorz Golebiowski

Thanks Grzegorz, nice trick!

--
You see things; and you say Why?
But I dream things that never were; and I say Why not?
-- George Bernard Shaw

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