Posted by AnrDaemon on 01/22/08 18:45

Greetings, mosesdinakaran@gmail.com.
In reply to Your message dated Tuesday, January 22, 2008, 17:41:34,

> Hi Thanks for the reply,

Hi and no thanks for the top-posting.

>> > Can any one explain how the rule is applied for the following Regular
>> > expression
>>
>> > $Str = 'the red king';
>>
>> > $Pattern = '/((red|white) (king|queen))/';
>>
>> $Pattern = '/( # start of capture no.1
>> ( # start of capture no.2
>> red|white # either literal 'red' or literal 'white'
>> ) # end of capture no.2
>> \s # space in originial, any whitespace for this one
>> ( # start of capture no.3
>> king|queen # either literal 'king' or literal 'queen'
>> ) # end of capture no.3
>> ) # end of capture no.1
>> /x';
>>
>> > Array
>> > (
>> > [0] => red king
>> > [1] => red king
>> > [2] => red
>> > [3] => king
>> > )
>>
>> Capture (1) is useless, as it will have exactly the same contents as the
>> whole match in (0), so lose the outside ( and ). If you don't need the
>> 'red' & 'king' in your match seperately, you can use (?: to create an
>> uncaptured subpattern.
>>
>> $Pattern = '/(?:red|white) (?:king|queen)/';

> Can you please tell me what I have under stood is right?

> After the start of capture no.2 the pattern starts with the word
> r is matched with t (the red king) -> NO Match
> r is matched with h -> No Match
> r is matched with e -> No Match
> .
> .
> .
> r is matched with r -> Match Found
> Now the engine moves to the Next character in the pattern ie e
> re is matched with re -> Match Found
> red is matched with red -> Match Found

Actually, the pattern is started to match "the red king" with "(red|white)"
(mean, two matches, one with "red" and one with "white") and advance by
character till string will be matched, "the red king" -> "he red king" ->
"e red king" -> "<space>red king" -> "red king" -> Match (red) found!
As far as we have this match in brackets, we saving the (red) as #1 match with
intention to return it as 1st substring.

> Since we have a | it comes out from the #first capture and now we
> have an empty space

> red(space) is matched with red(space)

Nop, as far as we have captured (red) we start the mew roundtrip.
"<space>king" will be matched against space in the pattern -> it matches, so
we saving it as #2 match (no return) and start third trip.

> red(space)k is matched with red(space)k
> .
> .
> red(space)king is matched with red(space)king -> Match Found

"king" will be matched against "(king|queen)" and it matches.
Let's save it as #3 match and remember to return it as 2nd subpattern
There is no more atoms in pattern, so let's look to see what we need else.
Oh, we have an outer brackets contains both subpatterns.
So we storing #1+#2+#3 as [1] and shifting numbers of #1 and #3.

> Now the value is returned and we get the first match

> array
> (
> [0] => the red king
> )

Yup, the [0] subpattern is the whole string where the match occured.

> Is the above mentioned procedure is correct?

> Also I was not able to find out how the other three values

> Array
> (
> [1] => red king
> [2] => red
> [3] => king
> )

> are matched.

As described.

Array
(
[0] => 'the red king' // Whole string where match occured
[1] => 'red king' // The 1st subpattern ((red)<space>(king))
[2] => 'red' // The 2nd subpattern (red)
[3] => 'king' // The 3rd subpattern (king)
)


--
Sincerely Yours, AnrDaemon <anrdaemon@freemail.ru>

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