Another way would be
if ($vaidateStr == preg_replace("/[^a-z0-9\-]/i", "", $vaidateStr)) {
/true, or valid...
}
the "i" after the last / makes it case insensitive. So it should match
upper and lower alpha characters.
> Ah yes, I also forgot the uppercase chars
>
> if ($vaidateStr == preg_replace("/[^a-zA-Z0-9\-]/", "", $vaidateStr)) {
> /true, or valid...
> }
>
> BTW also as Eric says, the carrot (^) inside the square brackets means
> "anything but" (negation), thus how this works..
>
> Ben
>
> On Wed, Jul 20, 2011 at 9:38 PM, Eric Cope <eric.cope@gmail.com
> <mailto:eric.cope@gmail.com>> wrote:
>
> Here is the regular expression:
>
> /^[a-zA-Z0-9\-]+$/
>
> the ^ requires the pattern match at the beginning of the string.
> The + requires at least one of these characters, but no upper limit.
> You can use * if you don't need the lower limit of 1.
> $ requires the patten match all the way to the end of the string.
> The dash requires escaping, hence the back slash.
>
> http://www.regular-expressions.info/anchors.html
>
> Thanks,
> Eric
>
> On Wed, Jul 20, 2011 at 8:34 PM, keith smith <klsmith2020@yahoo.com
> <mailto:klsmith2020@yahoo.com>> wrote:
>
>
> Hi,
>
> I have an input string that should only be lower or upper
> alphas, numbers and can contain a hyphen.
>
> I'm trying to figure out how to get PHP preg_match to verify the
> input string only contains these chars.
>
> I tried some thing like this and it always returns true if I
> have one or more of the specified chars, even if I have a char
> that is not specified.
>
> I need a true for any string that contains Alphas and/or numbers
> and/or one or more hyphens. I need it to return if the string
> contains anything else.
>
> preg_match("/[a-z0-9\+\-]/", $vaidateStr)
>
> I do not need to use preg_match, I'm looking for a simple solution.
>
> Any pointers are much appreciated.
>
> ------------------------
> Keith Smith
>
>
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> --
> ---
> Ben
>
> /print ''.join(map(lambda x: chr(x), ( (ord('a')-(3*5)),
> int(math.sqrt(math.pi*76)*5+2), int(math.ceil(math.e)*28),
> int(math.floor(math.e)*35), long(abs(4%3*35+3)*2))))/
>
>
>
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