Re: [Flashcoders] RegExp question
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 ^ and
the $ character.
You can also collapse down your expression to something a bit tighter.
This works (with the sample strings you gave):
^g(1|2|3)(e|w)\d{3}(-\d{2})?$
Which basically says match if:
The first character in the string is a g.
That's followed by a 1, 2 or 3.
Then a lower-case e or w (you could use (e|w|E|W) if you want to
handle uppercase too).
Then 3 digits
Then, possibly, a dash followed by two more digits.
Then the end of the string.
If any of that is false, the match won't happen.
By the way, a really good way to put together and test Regular
Expressions is to use Grant Skinner's online tool RegExr:
http://www.gskinner.com/RegExr/
Hope that helps (and gives you the results you want!)
Ian
On Thu, Nov 5, 2009 at 8:23 PM, Mendelsohn, Michael
michael.mendels...@fmglobal.com wrote:
Hi list...
I have some input text box with a go button that is enabled or disabled based
on an Event.CHANGE for the TextInput. As someone types in a string, a go
button becomes enabled or disabled, but I want the *entire* text in the text
box to be considered, not just a matched substring. Is this possible?
((TextField(e.target).text).match(/(((g|1|2|3)(E|W)\d{3})|((g|1|2|3)(E|W)\d{3}-?(\d{2})?))/i))?activateGoButton(true):activateGoButton(false);
For instance, the above code would enable the go button when you type the
text: g1e222- But, it shouldn't at that point, because it should be
expeciting two more numbers. Any tips are appreciated.
Thanks,
- Michael M.
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Re: [Flashcoders] RegExp question
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. ___ Flashcoders mailing list Flashcoders@chattyfig.figleaf.com http://chattyfig.figleaf.com/mailman/listinfo/flashcoders
RE: [Flashcoders] RegExp question
Thanks very much Ian and Henrik. The ^ and $ was exactly what I was looking for. Regards, - Michael M. ___ Flashcoders mailing list Flashcoders@chattyfig.figleaf.com http://chattyfig.figleaf.com/mailman/listinfo/flashcoders
Re: [Flashcoders] regexp question
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 (so var pattern:RegExp = /000/g;), it will only pick up the first 000 (and every multiple thereafter) instead of leaving the first 0 (which is expected behaviour but something i'd like to get around) How about positive lookaheads? /(000)(?=(?:000)*[^0])/g If you want this also to work for 1234567 = 1,234,567, you can replace every 0 in the pattern with \d and call the replace function with , \1 instead of ,000 This is untested, but should work... Let me know. regards, Claudius ___ Flashcoders mailing list Flashcoders@chattyfig.figleaf.com http://chattyfig.figleaf.com/mailman/listinfo/flashcoders ___ Flashcoders mailing list Flashcoders@chattyfig.figleaf.com http://chattyfig.figleaf.com/mailman/listinfo/flashcoders
Re: [Flashcoders] regexp question
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 = /(\d\d\d)(?=(?:\d\d\d)*[^\d])/g; sYear = sYear.replace(pattern, ,$1); //traces 1234567 if i drop the NOT part var sYear:String = 1234567; var pattern:RegExp = /(\d\d\d)(?=(?:\d\d\d)*)/g; sYear = sYear.replace(pattern, ,$1); //traces ,123,4567 This stuff is really new to me so i really appreciate the help thanks a 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 (so var pattern:RegExp = /000/g;), it will only pick up the first 000 (and every multiple thereafter) instead of leaving the first 0 (which is expected behaviour but something i'd like to get around) How about positive lookaheads? /(000)(?=(?:000)*[^0])/g If you want this also to work for 1234567 = 1,234,567, you can replace every 0 in the pattern with \d and call the replace function with , \1 instead of ,000 This is untested, but should work... Let me know. regards, Claudius ___ Flashcoders mailing list Flashcoders@chattyfig.figleaf.com http://chattyfig.figleaf.com/mailman/listinfo/flashcoders ___ Flashcoders mailing list Flashcoders@chattyfig.figleaf.com http://chattyfig.figleaf.com/mailman/listinfo/flashcoders
Re: [Flashcoders] regexp question
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 whttp://www.regular-expressions.info/quickstart.html and http://www.regular-expressions.info/tutorial.html Hope this will get you started, Sid On Jul 4, 2008, at 11:32 AM, Allandt Bik-Elliott (Receptacle) wrote: 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 = /(\d\d\d)(?=(?:\d\d\d)*[^\d])/g; sYear = sYear.replace(pattern, ,$1); //traces 1234567 if i drop the NOT part var sYear:String = 1234567; var pattern:RegExp = /(\d\d\d)(?=(?:\d\d\d)*)/g; sYear = sYear.replace(pattern, ,$1); //traces ,123,4567 This stuff is really new to me so i really appreciate the help thanks a 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 (so var pattern:RegExp = /000/g;), it will only pick up the first 000 (and every multiple thereafter) instead of leaving the first 0 (which is expected behaviour but something i'd like to get around) How about positive lookaheads? /(000)(?=(?:000)*[^0])/g If you want this also to work for 1234567 = 1,234,567, you can replace every 0 in the pattern with \d and call the replace function with , \1 instead of ,000 This is untested, but should work... Let me know. regards, Claudius ___ Flashcoders mailing list Flashcoders@chattyfig.figleaf.com http://chattyfig.figleaf.com/mailman/listinfo/flashcoders ___ Flashcoders mailing list Flashcoders@chattyfig.figleaf.com http://chattyfig.figleaf.com/mailman/listinfo/flashcoders ___ Flashcoders mailing list Flashcoders@chattyfig.figleaf.com http://chattyfig.figleaf.com/mailman/listinfo/flashcoders
Re: [Flashcoders] regexp question
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 RegEx on www.ilovejackdaniels.com and there is ofcourse alot on whttp://www.regular-expressions.info/ quickstart.html and http://www.regular-expressions.info/tutorial.html Hope this will get you started, Sid On Jul 4, 2008, at 11:32 AM, Allandt Bik-Elliott (Receptacle) wrote: 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 = /(\d\d\d)(?=(?:\d\d\d)*[^\d])/g; sYear = sYear.replace(pattern, ,$1); //traces 1234567 if i drop the NOT part var sYear:String = 1234567; var pattern:RegExp = /(\d\d\d)(?=(?:\d\d\d)*)/g; sYear = sYear.replace(pattern, ,$1); //traces ,123,4567 This stuff is really new to me so i really appreciate the help thanks a 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 (so var pattern:RegExp = /000/g;), it will only pick up the first 000 (and every multiple thereafter) instead of leaving the first 0 (which is expected behaviour but something i'd like to get around) How about positive lookaheads? /(000)(?=(?:000)*[^0])/g If you want this also to work for 1234567 = 1,234,567, you can replace every 0 in the pattern with \d and call the replace function with ,\1 instead of ,000 This is untested, but should work... Let me know. regards, Claudius ___ Flashcoders mailing list Flashcoders@chattyfig.figleaf.com http://chattyfig.figleaf.com/mailman/listinfo/flashcoders ___ Flashcoders mailing list Flashcoders@chattyfig.figleaf.com http://chattyfig.figleaf.com/mailman/listinfo/flashcoders ___ Flashcoders mailing list Flashcoders@chattyfig.figleaf.com http://chattyfig.figleaf.com/mailman/listinfo/flashcoders ___ Flashcoders mailing list Flashcoders@chattyfig.figleaf.com http://chattyfig.figleaf.com/mailman/listinfo/flashcoders
RE: [Flashcoders] regexp question
Hi, var sYear:String = 1234567; var pattern:RegExp = /(\d\d\d)(?=(?:\d\d\d)*[^\d])/g; sYear = sYear.replace(pattern, ,$1); //traces 1234567 That's because [^\d] expects a non-digit after the number Try this: var sYear:String = The year when all happened was 1234567 indeed // :) To get this to also work with just the year you may replace [^\d] with (?:[^\d]|$) which expects a non-digit or the end of the string regards Claudius ___ Flashcoders mailing list Flashcoders@chattyfig.figleaf.com http://chattyfig.figleaf.com/mailman/listinfo/flashcoders
RE: [Flashcoders] regexp question
To get this to also work with just the year you may replace [^\d] with (?:[^\d]|$) which expects a non-digit or the end of the string Or even better Replace [^\d] with \b which should also work. regards Claudius ___ Flashcoders mailing list Flashcoders@chattyfig.figleaf.com http://chattyfig.figleaf.com/mailman/listinfo/flashcoders
Re: [Flashcoders] regexp question
the \b boundary worked a treat - i'm just researching why it worked now thanks for all your help guys a On 4 Jul 2008, at 13:17, Claudius Ceteras wrote: To get this to also work with just the year you may replace [^\d] with (?:[^\d]|$) which expects a non-digit or the end of the string Or even better Replace [^\d] with \b which should also work. regards Claudius ___ Flashcoders mailing list Flashcoders@chattyfig.figleaf.com http://chattyfig.figleaf.com/mailman/listinfo/flashcoders ___ Flashcoders mailing list Flashcoders@chattyfig.figleaf.com http://chattyfig.figleaf.com/mailman/listinfo/flashcoders
RE: [Flashcoders] regexp question
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 at the end of the number, so because it only matches three digits groups which are followed by a number of digits which is divisible by 3, it doesnt find 123 in 1234567, but 234 (because it's followed by 3 digits followed by a word boundary) and 456 (because it's followed by 0 digits followed by a wod boundary) Btw, (?: ) is a non-capturing group, (?= ) is a positive lookahead. You can google for both... regards Claudius ___ Flashcoders mailing list Flashcoders@chattyfig.figleaf.com http://chattyfig.figleaf.com/mailman/listinfo/flashcoders
Re: [Flashcoders] regexp question
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 followed by positive lookahead: (?= ) 0, 3, 6, 9, ... Digits, followed by a word boundary (?:\d\d\d)*\b Word boundaries match at the end of the number, so because it only matches three digits groups which are followed by a number of digits which is divisible by 3, it doesnt find 123 in 1234567, but 234 (because it's followed by 3 digits followed by a word boundary) and 456 (because it's followed by 0 digits followed by a wod boundary) Btw, (?: ) is a non-capturing group, (?= ) is a positive lookahead. You can google for both... regards Claudius ___ Flashcoders mailing list Flashcoders@chattyfig.figleaf.com http://chattyfig.figleaf.com/mailman/listinfo/flashcoders ___ Flashcoders mailing list Flashcoders@chattyfig.figleaf.com http://chattyfig.figleaf.com/mailman/listinfo/flashcoders
RE: [Flashcoders] regexp question
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 leaving the first 0 (which is expected behaviour but something i'd like to get around) How about positive lookaheads? /(000)(?=(?:000)*[^0])/g If you want this also to work for 1234567 = 1,234,567, you can replace every 0 in the pattern with \d and call the replace function with ,\1 instead of ,000 This is untested, but should work... Let me know. regards, Claudius ___ Flashcoders mailing list Flashcoders@chattyfig.figleaf.com http://chattyfig.figleaf.com/mailman/listinfo/flashcoders
Re: [Flashcoders] regexp question
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, except for some reason you want to roll your own.
Cheers
Juan Pablo Califano
2008/7/2, Allandt Bik-Elliott (Receptacle) [EMAIL PROTECTED]:
hey
i am using regexp to inject commas into my years by searching for 000 and
replacing with ,000 like this
var pattern:RegExp = /000/;
sYear = sYear.replace(pattern, ,000);
however, this approach will bug every multiple of 10,000 as there are more
zeros than the pattern expects.
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 leaving the first 0 (which is
expected behaviour but something i'd like to get around)
would you use an if statement to count the length of the sYear string or is
there a better way?
at the moment, because i know that i only need years from 15,000 bc, i'm
doing this:
var sYear:String = String(nYear);
if (sYear.length 5)
{
if (sYear != -1)
{
var pattern:RegExp = /000/;
sYear = sYear.replace(pattern,
,000);
} else {
var pattern:RegExp = //;
sYear = sYear.replace(pattern,
0,000);
}
}
which is a bit hacky and limited
interested to hear your answer
a
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Re: [Flashcoders] regexp question
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 formatNumber method you can find in many other libraries
will
help you too, except for some reason you want to roll your own.
Cheers
Juan Pablo Califano
2008/7/2, Allandt Bik-Elliott (Receptacle)
[EMAIL PROTECTED]:
hey
i am using regexp to inject commas into my years by searching for
000 and
replacing with ,000 like this
var pattern:RegExp = /000/;
sYear = sYear.replace(pattern, ,000);
however, this approach will bug every multiple of 10,000 as there
are more
zeros than the pattern expects.
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 leaving the first 0
(which is
expected behaviour but something i'd like to get around)
would you use an if statement to count the length of the sYear
string or is
there a better way?
at the moment, because i know that i only need years from 15,000
bc, i'm
doing this:
var sYear:String = String(nYear);
if (sYear.length 5)
{
if (sYear != -1)
{
var pattern:RegExp = /000/;
sYear = sYear.replace(pattern,
,000);
} else {
var pattern:RegExp = //;
sYear = sYear.replace(pattern,
0,000);
}
}
which is a bit hacky and limited
interested to hear your answer
a
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