Hi
Im rather new to PHP (only been at it a week) so i hope someone can help me

I want to check a string to see if its a valid phonenumber, so it may contain only numbers and spaces AND it may also contain a single plus(+) sign for contry code, but only at the beginning of the number

this is what i have come up with so far:

First chekking the string only have numbers, spaces and +sign
if (preg_match("/^([0-9 \+]+)$/", $string)) {

Then chekking that ther is only +sign
if (preg_match_all("/[\+]/", $string)==1) {
$check2 = 'Landcode ok<br>';
}
else
{
$check2 = 'Landcode NOT ok<br>';
}
and its this check i cant get to work ??? no mater

Please try the following




if (preg_match("/^(\+)*([0-9]+)$/", $string)) {

echo "Okay";

} else {

echo "not a land number";
}

if (preg_match("/^\+?([\d\s]+)$/", $string)) {

Thanks :)

As i see it i wasn't totaly wrong
But that looks mutch easyer

Only thing missing was the space but got that with [0-9 ]

now i haw to find out hov/why that works ?

i gues (if i have read my tutorials corect) that the ^(\+) meens look for a + ONLY in the beginning
the * is like a + but may only ocur once
[0-9 ]+ is looking for number and spaces and the space meens the may ocur more than once
and the $ meen its shal keep on looking until the end

??

Thanks ... thant workd to

almost the same .. but with \d for digits and \s for spaces right ??

but that ? and no () around \+ ???

Thanks ... thant workd to

almost the same .. but with \d for digits and \s for spaces right ??

but that ? and no () around \+ ???

You aren't capturing the plus sign? No need for the capturing parantheses, if so. The ? in the expression states that the preceding character may or may not exist, but if it does it will match one, and only one, plus sign, whereas the * in the first example posted will match none of or any number of + at the beginning of the string. Try it with several ++ at the beginning of the string and you'll see what I mean. The \d and \s, as you correctly surmise, are merely shorthand.


Edit: In fact, you aren't capturing anything with that expression, so you can remove all parantheses:


if (preg_match("/^\+?[\d\s]+$/", $string)) {


If you want to use it as a check and capture expression, use:


if (preg_match("/^(\+?[\d\s]+)$/", $string, $matches)) {

Your right .. didnt test with 2+ signs ind the begining

BTW.
Is it posible to "revers" that line so the if state only is tru if it dont match

Thinking loosing the else state
somthing like:
ifnot (preg_match("/^\+*([\d\s]+)$/", $string)) {
echo ('WRONG');
}

You aren't capturing the plus sign? No need for the capturing parantheses, ......

Edit: In fact, you aren't capturing anything with that expression, so you can remove all parantheses:


Im not shure i know what is ment by capturing
dictionary says To take captive,seize, To gain possession or contro ??

Sorry my english aint that good (Im a little dyslexie)

For a direct non match, use:


if (!preg_match("/^\+?[\d\s]+$/", $string)) {

Im not shure i know what is ment by capturing

If you want to capture, (have the telephone number available for later use in your script), the capturing expression I posted would make the phone number available via the $matches[1] var.

funny tryd that befor asking... but cud not get it to work .. but now it dos ..

May be becaus it 8am and i have been working on this all night :-P

Ahhh ... ok

I am going to use it later ... alogn with other entrys like name and adress
Ill look into that $matches[]

Ahhh ... ok

I am going to use it later ... alogn with other entrys like name and adress
Ill look into that $matches[]

It's a case of either/either, to be honest. The $string var that you're checking against would, (provided it passes the preg_match check), already contain a valid phone number. The $matches var from the preg_match would only come into it's own if you were literally extracting the phone number from a longer string. The fact that you're tying the expression to the start and end of the string with ^ and $ respectively means that you're safe just using $string as your number, once it has passed the validation check.

This is why I love regex buddy
It can explain the whole regex for you. This is the explanation for MattF's regex
Regex info

^\+?([\d\s]+)$

Assert position at the beginning of the string ^
Match the character “+” literally \+?
Between zero and one times, as many times as possible, giving back as needed (greedy) ?
Match the regular expression below and capture its match into backreference number 1 ([\d\s]+)
Match a single character present in the list below [\d\s]+
Between one and unlimited times, as many times as possible, giving back as needed (greedy) +
A single digit 0..9 \d
A whitespace character (spaces, tabs, line breaks, etc.) \s
Assert position at the end of the string (or before the line break at the end of the string, if any) $

Hope that helps explain the whole process for you :)

Thanks for all your help.

Learning php is mutch easyer when peapol are so helpsful explaining :-)

This is why I love regex buddy
It can explain the whole regex for you.

Sadly regex buddy aint for mac then :(

Maby i can find somthing like it for my mac

Nope it's not free either, but it's a damn good tool

Nope it's not free either, but it's a damn good tool

Well not many god things ar free :)

I dit a little google for a mac substitute
found a few here
http://osx.iusethis.com/search?q=regex