Hi Michael,
Firstly, I'm not quite sure your expression is right - it says g
_or_ 1 or 2 or 3 as the first character, whereas your sample starts
with a g then a 1.
But anyway - what you need to do is to test for the beginning and
end of the string. In regular expressions, you do this with the
Mendelsohn, Michael wrote:
I want the *entire* text in the text box to be considered, not just a matched
substring. Is this possible?
Use ^ to lock to the begining of the string and $ to lock to the end of
the string, use both and it will not be allowed anything other than the
expression.
Thanks very much Ian and Henrik. The ^ and $ was exactly what I was looking
for.
Regards,
- Michael M.
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wow - i have no idea what that means at all
time to hit the books - thanks :)
On 4 Jul 2008, at 01:09, Claudius Ceteras wrote:
Hi,
is there a way of counting back from the end of the number and
inserting the comma (even without a regular expression)? if i
use the
g modifier in the regexp
hi again
i've been trying different things and it seems that the [^0] or [^\d]
is stopping it working. (I needed to use $1 rather than \1 to
reference the first group in the String.replace statement)
here is what i've got so far
var sYear:String = 1234567;
var pattern:RegExp =
Hi Allandt,
Have you found this tool already?
http://www.gskinner.com/blog/archives/2008/03/regexr_free_onl.html
It allows you to test your regex pattern expecially for AS
And you can find a cheatsheat on RegEx on www.ilovejackdaniels.com and
there is ofcourse alot on
hey that's great sid - thanks
a
On 4 Jul 2008, at 12:26, Sidney de Koning wrote:
Hi Allandt,
Have you found this tool already? http://www.gskinner.com/blog/
archives/2008/03/regexr_free_onl.html
It allows you to test your regex pattern expecially for AS
And you can find a cheatsheat on
Hi,
var sYear:String = 1234567;
var pattern:RegExp = /(\d\d\d)(?=(?:\d\d\d)*[^\d])/g;
sYear = sYear.replace(pattern, ,$1);
//traces 1234567
That's because [^\d] expects a non-digit after the number
Try this:
var sYear:String = The year when all happened was 1234567 indeed // :)
To get
To get this to also work with just the year you may replace [^\d] with
(?:[^\d]|$) which expects a non-digit or the end of the string
Or even better Replace [^\d] with \b which should also work.
regards
Claudius
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the \b boundary worked a treat - i'm just researching why it worked now
thanks for all your help guys
a
On 4 Jul 2008, at 13:17, Claudius Ceteras wrote:
To get this to also work with just the year you may replace [^\d]
with
(?:[^\d]|$) which expects a non-digit or the end of the string
the \b boundary worked a treat - i'm just researching why it
worked now
/(\d\d\d)(?=(?:\d\d\d)*\b)/g
Find all groups of three digits (\d\d\d) , which are followed by
positive lookahead: (?= )
0, 3, 6, 9, ... Digits, followed by a word boundary (?:\d\d\d)*\b
Word boundaries match
wow - that's really helpful - thanks a lot for your time claudius
best
a
On 4 Jul 2008, at 14:56, Claudius Ceteras wrote:
the \b boundary worked a treat - i'm just researching why it
worked now
/(\d\d\d)(?=(?:\d\d\d)*\b)/g
Find all groups of three digits (\d\d\d) , which are
Hi,
is there a way of counting back from the end of the number and
inserting the comma (even without a regular expression)? if i
use the
g modifier in the regexp (so var pattern:RegExp = /000/g;), it will
only pick up the first 000 (and every multiple thereafter)
instead of
Hi,
You can check out ascb (ActionScript Cook Book), a library with some useful
functions. In this case, the class NumberFormat, and the method format may
do the job.
http://www.rightactionscript.com/ascb/
Almost any formatNumber method you can find in many other libraries will
help you too,
thanks a lot :)
On 2 Jul 2008, at 15:24, Juan Pablo Califano wrote:
Hi,
You can check out ascb (ActionScript Cook Book), a library with
some useful
functions. In this case, the class NumberFormat, and the method
format may
do the job.
http://www.rightactionscript.com/ascb/
Almost any
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