regex - Using if and else in regular expression -

i'm having difficulty trying understand particular regular expression (it used check user input phone number) :

^((\+\d{1,3}(-| )?\(?\d\)?(-| )?\d{1,3})|(\(?\d{2,3}\)?))(-| )?(\d{1,4})(-| )?(\d{6})(( x| ext)\d{1,5}){0,1}$ 

i read "?()" used if condition in regular expression, still not clear me logic behind regular expression , kind of input accepted , rejected it.

thanks

firstly, in regexp, ?() not conditional. ? matches character (group) left of 0 or 1 times , () starts capture group nothing in it?... no conditionals i'm afraid :) closest might (a|b) matches either a or b...

the regexp little difficult read, so

^((\+\d{1,3}(-| )?\(?\d\)?(-| )?\d{1,3})|(\(?\d{2,3}\)?))(-| )?(\d{1,4})(-| )?(\d{6})(( x| ext)\d{1,5}){0,1}$ 

try regexper.com, type in regexp , draw state diagram...

using tabbing break expression:

^(     (\+\d{1,3}(-| )?     \(?\d\)?(-| )?     \d{1,3})   |(     \(?\d{2,3}\)?    ) )  (-| )?(\d{1,4}) (-| )?(\d{6}) (     ( x| ext)\d{1,5} ){0,1}$ 

(note makes spaces hard read we'll go through referencing original)

^ matches start of line

the next group ((\+\d{1,3}(-| )?\(?\d\)?(-| )?\d{1,3})|(\(?\d{2,3}\)?))

this has 2 parts: (x|y), x=(\+\d{1,3}(-| )?\(?\d\)?(-| )?\d{1,3}) , y=(\(?\d{2,3}\)?). match either x or y...

breaking down x=(\+\d{1,3}(-| )?\(?\d\)?(-| )?\d{1,3}):

1. the outer () capture, strip these...

2. \+ matches literal plus sign. note has escaped \ because + meta character meaning "match 1 or more of previous".

3. \d{1,3} matches decimal digit eiter 1, 2 or 3 times no more or less

4. (-| )? matches either - or (space) 0 or 1 times. ? wht specifies 0 or 1 times.

5. \(?\d\) matches literal '(' (notice escape) 0 or 1 times. decimal digit, literal )

6. (-| )? we've seen before (matches either - or (space) 0 or 1 times. ? wht specifies 0 or 1 times.)

7. \d{1,3} we've seen before (matches decimal digit eiter 1, 2 or 3 times no more or less)

so can x matches (and captures - that's wat outer () doing) string starts plus, has 1 3 digits possibly space or hyphen, digit inside brackets, possibly space or hyphen , 1 3 digits. captured first capture group... phew!

breaking down y=(\(?\d{2,3}\)?):

1. the outer () capture string these...

2. \(? matches literal ( 0 or 1 times.

3. \d{2,3} matches digit 2 or 3 times

4. \)? matches literal ) 0 or 1 times

so can y matches 2 or 3 digit number, possibly surrounded brackets. captures first capture group. jeez!

now have x , y can see first chunk of regexp matches (brain melting!).

the first chunk, call chunk1 matches , captures either

1. any string starts plus, has 1 3 digits possibly space or hyphen, digit inside brackets, possibly space or hyphen , 1 3 digits or

2. any 2 or 3 digit number, possibly surrounded brackets

continuing... (-| )? we've seen before (matches either - or (space) 0 or 1 times. ? wht specifies 0 or 1 times.)

(\d{1,4}) matches string of digit characters 1,2,3 or 4 digits in length. forms second capture group.

(-| )? we've seen before (matches either - or (space) 0 or 1 times. ? wht specifies 0 or 1 times.)

(\d{6}) matches string of 6 digits

so here matching string possible space or hypen, 1 4 numbers, possible space or hyphen , 6 numbers. call chunk2

so far have matched string consistiing of chunk1 followed chunk2...

this concludes main bit of phone number, rest appears handle extensions...

the next bit (( x| ext)\d{1,5}){0,1}. lets break down little.

1. the surrounding brackets capture group.

2. ( x| ext) matches either of 2 literal strings ' x' or ' ext' - note beginning space.

3. \d{1,5} matches digit 1,2,3,4 or 5 times.

4. {0,1} matches capture group 0 or 1 times... i.e. phone number not need have extension

finally $ matches end of line.

hopefully has broken down string enough work through :)