Regex. Split or Split

microsoft.public.dotnet.languages.vb

13 Apr 2005 1:25 PM

lgbjr

Hi All,

I'm trying to split a string on every character. The string happens to be a
representation of a hex number. So, my regex expression is ([A-F,0-9]).
Seems simple, but for some reason, I'm not getting the results I expect.

Dim SA as string()
Dim S as string

S="FBE"
SA=RegularExpressions.Regex.Split(S,"([A-F,0-9])")

I expect that SA will contain 3 elements: SA(0)="F", SA(1)="B", SA(2)="E",
but, what I'm getting is 7 elements: SA(0)="", SA(1)="F", SA(2)="",
SA(3)="B", SA(4)="", SA(5)="E", SA(6)="".

If I change the expression to [A-F,0-9] (no parentheses), I get: SA(0)="",
SA(1)="", SA(2)="", SA(3)="".

Just for my own sanity, I've checked the pattern in Expresso and it returns
what I would expect.

I suppose I should mention I'm using VB.NET 2005 (just in case there's a
known issue with Regex in 2005).

TIA
Lee

13 Apr 2005 1:47 PM

Jay B. Harlow [MVP - Outlook]

Lee,

| I expect that SA will contain 3 elements: SA(0)="F", SA(1)="B", SA(2)="E",
| but, what I'm getting is 7 elements: SA(0)="", SA(1)="F", SA(2)="",
| SA(3)="B", SA(4)="", SA(5)="E", SA(6)="".

I would expect it to contain 4 elements, SA(0)="", SA(1)="", SA(2)="", and
SA(3)="", as your string only contains delimiters. RegEx.Split returns the
strings between the delimiters, unless you use capturing groups (the
parenthesis in your expression) in which case it returns both the strings
between the delimiters & the delimiters.

The pattern "[A-F,0-9]" returns the 4 that I expect. The capturing groups in
your expression is causing RegEx to return the 4 strings between the
delimiters, plus the 3 delimiters, ergo 7 values.

It sounds like you really want to return the list of matches, rather then
the stuff between the delimiters... Try RegEx.Matches, something like:

        Dim input As String = "FBE"

        Const pattern As String = "([A-F,0-9])"
        Static parser As New Regex(pattern)

        For Each match As Match In parser.Matches(input)
            Debug.WriteLine(match.Value)
        Next

Hope this helps
Jay

Show quoteHide quote

"lgbjr" <lgbjr@online.nospam> wrote in message
news:OlvF7wCQFHA.2356@TK2MSFTNGP14.phx.gbl...
| Hi All,
|
| I'm trying to split a string on every character. The string happens to be
a
| representation of a hex number. So, my regex expression is ([A-F,0-9]).
| Seems simple, but for some reason, I'm not getting the results I expect.
|
| Dim SA as string()
| Dim S as string
|
| S="FBE"
| SA=RegularExpressions.Regex.Split(S,"([A-F,0-9])")
|
| I expect that SA will contain 3 elements: SA(0)="F", SA(1)="B", SA(2)="E",
| but, what I'm getting is 7 elements: SA(0)="", SA(1)="F", SA(2)="",
| SA(3)="B", SA(4)="", SA(5)="E", SA(6)="".
|
| If I change the expression to [A-F,0-9] (no parentheses), I get: SA(0)="",
| SA(1)="", SA(2)="", SA(3)="".
|
| Just for my own sanity, I've checked the pattern in Expresso and it
returns
| what I would expect.
|
| I suppose I should mention I'm using VB.NET 2005 (just in case there's a
| known issue with Regex in 2005).
|
| TIA
| Lee
|
|

13 Apr 2005 2:13 PM

lgbjr

Thanks Jay,

I'm an idiot!! an hour or so ago, I was working on a split (really a split),
and just continued with split when I should have been using Matches. LOL! Of
course, Expresso was giving me the result I expected, because it wasn't
trying to split the string!

thanks for pointing out what should have been an obvious mistake.

Lee

Show quoteHide quote

"Jay B. Harlow [MVP - Outlook]" <Jay_Harlow_***@msn.com> wrote in message
news:%23JuWn9CQFHA.3144@tk2msftngp13.phx.gbl...
> Lee,
> | I expect that SA will contain 3 elements: SA(0)="F", SA(1)="B",
> SA(2)="E",
> | but, what I'm getting is 7 elements: SA(0)="", SA(1)="F", SA(2)="",
> | SA(3)="B", SA(4)="", SA(5)="E", SA(6)="".
> I would expect it to contain 4 elements, SA(0)="", SA(1)="", SA(2)="", and
> SA(3)="", as your string only contains delimiters. RegEx.Split returns the
> strings between the delimiters, unless you use capturing groups (the
> parenthesis in your expression) in which case it returns both the strings
> between the delimiters & the delimiters.
>
> The pattern "[A-F,0-9]" returns the 4 that I expect. The capturing groups
> in
> your expression is causing RegEx to return the 4 strings between the
> delimiters, plus the 3 delimiters, ergo 7 values.
>
>
> It sounds like you really want to return the list of matches, rather then
> the stuff between the delimiters... Try RegEx.Matches, something like:
>
>        Dim input As String = "FBE"
>
>        Const pattern As String = "([A-F,0-9])"
>        Static parser As New Regex(pattern)
>
>        For Each match As Match In parser.Matches(input)
>            Debug.WriteLine(match.Value)
>        Next
>
> Hope this helps
> Jay
>
>
>
> "lgbjr" <lgbjr@online.nospam> wrote in message
> news:OlvF7wCQFHA.2356@TK2MSFTNGP14.phx.gbl...
> | Hi All,
> |
> | I'm trying to split a string on every character. The string happens to
> be
> a
> | representation of a hex number. So, my regex expression is ([A-F,0-9]).
> | Seems simple, but for some reason, I'm not getting the results I expect.
> |
> | Dim SA as string()
> | Dim S as string
> |
> | S="FBE"
> | SA=RegularExpressions.Regex.Split(S,"([A-F,0-9])")
> |
> | I expect that SA will contain 3 elements: SA(0)="F", SA(1)="B",
> SA(2)="E",
> | but, what I'm getting is 7 elements: SA(0)="", SA(1)="F", SA(2)="",
> | SA(3)="B", SA(4)="", SA(5)="E", SA(6)="".
> |
> | If I change the expression to [A-F,0-9] (no parentheses), I get:
> SA(0)="",
> | SA(1)="", SA(2)="", SA(3)="".
> |
> | Just for my own sanity, I've checked the pattern in Expresso and it
> returns
> | what I would expect.
> |
> | I suppose I should mention I'm using VB.NET 2005 (just in case there's a
> | known issue with Regex in 2005).
> |
> | TIA
> | Lee
> |
> |
>
>

13 Apr 2005 3:49 PM

Larry Lard

By the way, the comma in your regex pattern is not part of the syntax
of [] (ie what you actually mean is [A-F0-9] - at the moment you would
match , as a hex digit)

lgbjr wrote:
Show quoteHide quote

> Thanks Jay,
>
> I'm an idiot!! an hour or so ago, I was working on a split (really a
split),
> and just continued with split when I should have been using Matches.
LOL! Of
> course, Expresso was giving me the result I expected, because it
wasn't
> trying to split the string!
>
> thanks for pointing out what should have been an obvious mistake.
>
> Lee
>
> "Jay B. Harlow [MVP - Outlook]" <Jay_Harlow_***@msn.com> wrote in
message
> news:%23JuWn9CQFHA.3144@tk2msftngp13.phx.gbl...
> > Lee,
> > | I expect that SA will contain 3 elements: SA(0)="F", SA(1)="B",
> > SA(2)="E",
> > | but, what I'm getting is 7 elements: SA(0)="", SA(1)="F",
SA(2)="",
> > | SA(3)="B", SA(4)="", SA(5)="E", SA(6)="".
> > I would expect it to contain 4 elements, SA(0)="", SA(1)="",

SA(2)="", and
Show quoteHide quote

> > SA(3)="", as your string only contains delimiters. RegEx.Split
returns the
> > strings between the delimiters, unless you use capturing groups
(the
> > parenthesis in your expression) in which case it returns both the
strings
> > between the delimiters & the delimiters.
> >
> > The pattern "[A-F,0-9]" returns the 4 that I expect. The capturing
groups
> > in
> > your expression is causing RegEx to return the 4 strings between
the
> > delimiters, plus the 3 delimiters, ergo 7 values.
> >
> >
> > It sounds like you really want to return the list of matches,
rather then
> > the stuff between the delimiters... Try RegEx.Matches, something
like:
> >
> >        Dim input As String = "FBE"
> >
> >        Const pattern As String = "([A-F,0-9])"
> >        Static parser As New Regex(pattern)
> >
> >        For Each match As Match In parser.Matches(input)
> >            Debug.WriteLine(match.Value)
> >        Next
> >
> > Hope this helps
> > Jay
> >
> >
> >
> > "lgbjr" <lgbjr@online.nospam> wrote in message
> > news:OlvF7wCQFHA.2356@TK2MSFTNGP14.phx.gbl...
> > | Hi All,
> > |
> > | I'm trying to split a string on every character. The string
happens to
> > be
> > a
> > | representation of a hex number. So, my regex expression is

([A-F,0-9]).
Show quoteHide quote

there's a
Show quoteHide quote

> > | known issue with Regex in 2005).
> > |
> > | TIA
> > | Lee
> > |
> > |
> >
> >

14 Apr 2005 1:18 AM

Dennis

Why can't you use the .ToCharArray method of Strings?

Show quoteHide quote

"lgbjr" wrote:

> Thanks Jay,
>
> I'm an idiot!! an hour or so ago, I was working on a split (really a split),
> and just continued with split when I should have been using Matches. LOL! Of
> course, Expresso was giving me the result I expected, because it wasn't
> trying to split the string!
>
> thanks for pointing out what should have been an obvious mistake.
>
> Lee
>
> "Jay B. Harlow [MVP - Outlook]" <Jay_Harlow_***@msn.com> wrote in message
> news:%23JuWn9CQFHA.3144@tk2msftngp13.phx.gbl...
> > Lee,
> > | I expect that SA will contain 3 elements: SA(0)="F", SA(1)="B",
> > SA(2)="E",
> > | but, what I'm getting is 7 elements: SA(0)="", SA(1)="F", SA(2)="",
> > | SA(3)="B", SA(4)="", SA(5)="E", SA(6)="".
> > I would expect it to contain 4 elements, SA(0)="", SA(1)="", SA(2)="", and
> > SA(3)="", as your string only contains delimiters. RegEx.Split returns the
> > strings between the delimiters, unless you use capturing groups (the
> > parenthesis in your expression) in which case it returns both the strings
> > between the delimiters & the delimiters.
> >
> > The pattern "[A-F,0-9]" returns the 4 that I expect. The capturing groups
> > in
> > your expression is causing RegEx to return the 4 strings between the
> > delimiters, plus the 3 delimiters, ergo 7 values.
> >
> >
> > It sounds like you really want to return the list of matches, rather then
> > the stuff between the delimiters... Try RegEx.Matches, something like:
> >
> >        Dim input As String = "FBE"
> >
> >        Const pattern As String = "([A-F,0-9])"
> >        Static parser As New Regex(pattern)
> >
> >        For Each match As Match In parser.Matches(input)
> >            Debug.WriteLine(match.Value)
> >        Next
> >
> > Hope this helps
> > Jay
> >
> >
> >
> > "lgbjr" <lgbjr@online.nospam> wrote in message
> > news:OlvF7wCQFHA.2356@TK2MSFTNGP14.phx.gbl...
> > | Hi All,
> > |
> > | I'm trying to split a string on every character. The string happens to
> > be
> > a
> > | representation of a hex number. So, my regex expression is ([A-F,0-9]).
> > | Seems simple, but for some reason, I'm not getting the results I expect.
> > |
> > | Dim SA as string()
> > | Dim S as string
> > |
> > | S="FBE"
> > | SA=RegularExpressions.Regex.Split(S,"([A-F,0-9])")
> > |
> > | I expect that SA will contain 3 elements: SA(0)="F", SA(1)="B",
> > SA(2)="E",
> > | but, what I'm getting is 7 elements: SA(0)="", SA(1)="F", SA(2)="",
> > | SA(3)="B", SA(4)="", SA(5)="E", SA(6)="".
> > |
> > | If I change the expression to [A-F,0-9] (no parentheses), I get:
> > SA(0)="",
> > | SA(1)="", SA(2)="", SA(3)="".
> > |
> > | Just for my own sanity, I've checked the pattern in Expresso and it
> > returns
> > | what I would expect.
> > |
> > | I suppose I should mention I'm using VB.NET 2005 (just in case there's a
> > | known issue with Regex in 2005).
> > |
> > | TIA
> > | Lee
> > |
> > |
> >
> >
>
>
>

14 Apr 2005 1:42 AM

lgbjr

Dennis,

Previously, that is exactly what I was doing:

Dim SA as Array
Dim S as String
S="FBE"
SA=S.ToCharArray

This works fine. but, I'm trying to use Options Strict now, which means I
can't use

Dim SA as Array

I have to use

Dim SA as String()

And 1-dimensional array of Char can not be converted to 1-dimensional array
of String. So, I decided to use a Regex Match to convert the string to a
string array.

LOL! As I was typing this, I just realized I can do Dim SA as Char(), then
use .ToCharArray!!

Thanks!!

Lee

Show quoteHide quote

"Dennis" <Den***@discussions.microsoft.com> wrote in message
news:1D85F3C7-FC03-4B75-BA56-109445A9BF92@microsoft.com...
> Why can't you use the .ToCharArray method of Strings?
>
> "lgbjr" wrote:
>
>> Thanks Jay,
>>
>> I'm an idiot!! an hour or so ago, I was working on a split (really a
>> split),
>> and just continued with split when I should have been using Matches. LOL!
>> Of
>> course, Expresso was giving me the result I expected, because it wasn't
>> trying to split the string!
>>
>> thanks for pointing out what should have been an obvious mistake.
>>
>> Lee
>>
>> "Jay B. Harlow [MVP - Outlook]" <Jay_Harlow_***@msn.com> wrote in message
>> news:%23JuWn9CQFHA.3144@tk2msftngp13.phx.gbl...
>> > Lee,
>> > | I expect that SA will contain 3 elements: SA(0)="F", SA(1)="B",
>> > SA(2)="E",
>> > | but, what I'm getting is 7 elements: SA(0)="", SA(1)="F", SA(2)="",
>> > | SA(3)="B", SA(4)="", SA(5)="E", SA(6)="".
>> > I would expect it to contain 4 elements, SA(0)="", SA(1)="", SA(2)="",
>> > and
>> > SA(3)="", as your string only contains delimiters. RegEx.Split returns
>> > the
>> > strings between the delimiters, unless you use capturing groups (the
>> > parenthesis in your expression) in which case it returns both the
>> > strings
>> > between the delimiters & the delimiters.
>> >
>> > The pattern "[A-F,0-9]" returns the 4 that I expect. The capturing
>> > groups
>> > in
>> > your expression is causing RegEx to return the 4 strings between the
>> > delimiters, plus the 3 delimiters, ergo 7 values.
>> >
>> >
>> > It sounds like you really want to return the list of matches, rather
>> > then
>> > the stuff between the delimiters... Try RegEx.Matches, something like:
>> >
>> >        Dim input As String = "FBE"
>> >
>> >        Const pattern As String = "([A-F,0-9])"
>> >        Static parser As New Regex(pattern)
>> >
>> >        For Each match As Match In parser.Matches(input)
>> >            Debug.WriteLine(match.Value)
>> >        Next
>> >
>> > Hope this helps
>> > Jay
>> >
>> >
>> >
>> > "lgbjr" <lgbjr@online.nospam> wrote in message
>> > news:OlvF7wCQFHA.2356@TK2MSFTNGP14.phx.gbl...
>> > | Hi All,
>> > |
>> > | I'm trying to split a string on every character. The string happens
>> > to
>> > be
>> > a
>> > | representation of a hex number. So, my regex expression is
>> > ([A-F,0-9]).
>> > | Seems simple, but for some reason, I'm not getting the results I
>> > expect.
>> > |
>> > | Dim SA as string()
>> > | Dim S as string
>> > |
>> > | S="FBE"
>> > | SA=RegularExpressions.Regex.Split(S,"([A-F,0-9])")
>> > |
>> > | I expect that SA will contain 3 elements: SA(0)="F", SA(1)="B",
>> > SA(2)="E",
>> > | but, what I'm getting is 7 elements: SA(0)="", SA(1)="F", SA(2)="",
>> > | SA(3)="B", SA(4)="", SA(5)="E", SA(6)="".
>> > |
>> > | If I change the expression to [A-F,0-9] (no parentheses), I get:
>> > SA(0)="",
>> > | SA(1)="", SA(2)="", SA(3)="".
>> > |
>> > | Just for my own sanity, I've checked the pattern in Expresso and it
>> > returns
>> > | what I would expect.
>> > |
>> > | I suppose I should mention I'm using VB.NET 2005 (just in case
>> > there's a
>> > | known issue with Regex in 2005).
>> > |
>> > | TIA
>> > | Lee
>> > |
>> > |
>> >
>> >
>>
>>
>>

14 Apr 2005 8:23 PM

Jay B. Harlow [MVP - Outlook]

lager,

| LOL! As I was typing this, I just realized I can do Dim SA as Char(), then
| use .ToCharArray!!

Also depending on what you are doing with the Char(), you may be able to
simply use the String.Chars property.

Something like:

        Dim S As String
        S = "FBE"
        For index As Integer = 0 To S.Length - 1
            Dim ch As Char = S.Chars(index)
            If ch = ","c Then
                ' do something interesting with the comma
            End If
        Next

Hope this helps
Jay

Show quoteHide quote

"lgbjr" <lgbjr@online.nospam> wrote in message
news:%237y63MJQFHA.2580@TK2MSFTNGP10.phx.gbl...
| Dennis,
|
| Previously, that is exactly what I was doing:
|
| Dim SA as Array
| Dim S as String
| S="FBE"
| SA=S.ToCharArray
|
| This works fine. but, I'm trying to use Options Strict now, which means I
| can't use
|
| Dim SA as Array
|
| I have to use
|
| Dim SA as String()
|
| And 1-dimensional array of Char can not be converted to 1-dimensional
array
| of String. So, I decided to use a Regex Match to convert the string to a
| string array.
|
| LOL! As I was typing this, I just realized I can do Dim SA as Char(), then
| use .ToCharArray!!
|
| Thanks!!
|
| Lee
|
| "Dennis" <Den***@discussions.microsoft.com> wrote in message
| news:1D85F3C7-FC03-4B75-BA56-109445A9BF92@microsoft.com...
| > Why can't you use the .ToCharArray method of Strings?
| >
| > "lgbjr" wrote:
| >
| >> Thanks Jay,
| >>
| >> I'm an idiot!! an hour or so ago, I was working on a split (really a
| >> split),
| >> and just continued with split when I should have been using Matches.
LOL!
| >> Of
| >> course, Expresso was giving me the result I expected, because it wasn't
| >> trying to split the string!
| >>
| >> thanks for pointing out what should have been an obvious mistake.
| >>
| >> Lee
| >>
| >> "Jay B. Harlow [MVP - Outlook]" <Jay_Harlow_***@msn.com> wrote in
message
| >> news:%23JuWn9CQFHA.3144@tk2msftngp13.phx.gbl...
| >> > Lee,
| >> > | I expect that SA will contain 3 elements: SA(0)="F", SA(1)="B",
| >> > SA(2)="E",
| >> > | but, what I'm getting is 7 elements: SA(0)="", SA(1)="F", SA(2)="",
| >> > | SA(3)="B", SA(4)="", SA(5)="E", SA(6)="".
| >> > I would expect it to contain 4 elements, SA(0)="", SA(1)="",
SA(2)="",
| >> > and
| >> > SA(3)="", as your string only contains delimiters. RegEx.Split
returns
| >> > the
| >> > strings between the delimiters, unless you use capturing groups (the
| >> > parenthesis in your expression) in which case it returns both the
| >> > strings
| >> > between the delimiters & the delimiters.
| >> >
| >> > The pattern "[A-F,0-9]" returns the 4 that I expect. The capturing
| >> > groups
| >> > in
| >> > your expression is causing RegEx to return the 4 strings between the
| >> > delimiters, plus the 3 delimiters, ergo 7 values.
| >> >
| >> >
| >> > It sounds like you really want to return the list of matches, rather
| >> > then
| >> > the stuff between the delimiters... Try RegEx.Matches, something
like:
| >> >
| >> >        Dim input As String = "FBE"
| >> >
| >> >        Const pattern As String = "([A-F,0-9])"
| >> >        Static parser As New Regex(pattern)
| >> >
| >> >        For Each match As Match In parser.Matches(input)
| >> >            Debug.WriteLine(match.Value)
| >> >        Next
| >> >
| >> > Hope this helps
| >> > Jay
| >> >
| >> >
| >> >
| >> > "lgbjr" <lgbjr@online.nospam> wrote in message
| >> > news:OlvF7wCQFHA.2356@TK2MSFTNGP14.phx.gbl...
| >> > | Hi All,
| >> > |
| >> > | I'm trying to split a string on every character. The string happens
| >> > to
| >> > be
| >> > a
| >> > | representation of a hex number. So, my regex expression is
| >> > ([A-F,0-9]).
| >> > | Seems simple, but for some reason, I'm not getting the results I
| >> > expect.
| >> > |
| >> > | Dim SA as string()
| >> > | Dim S as string
| >> > |
| >> > | S="FBE"
| >> > | SA=RegularExpressions.Regex.Split(S,"([A-F,0-9])")
| >> > |
| >> > | I expect that SA will contain 3 elements: SA(0)="F", SA(1)="B",
| >> > SA(2)="E",
| >> > | but, what I'm getting is 7 elements: SA(0)="", SA(1)="F", SA(2)="",
| >> > | SA(3)="B", SA(4)="", SA(5)="E", SA(6)="".
| >> > |
| >> > | If I change the expression to [A-F,0-9] (no parentheses), I get:
| >> > SA(0)="",
| >> > | SA(1)="", SA(2)="", SA(3)="".
| >> > |
| >> > | Just for my own sanity, I've checked the pattern in Expresso and it
| >> > returns
| >> > | what I would expect.
| >> > |
| >> > | I suppose I should mention I'm using VB.NET 2005 (just in case
| >> > there's a
| >> > | known issue with Regex in 2005).
| >> > |
| >> > | TIA
| >> > | Lee
| >> > |
| >> > |
| >> >
| >> >
| >>
| >>
| >>
|
|

14 Apr 2005 11:11 PM

Dennis

My Point exactly..why use Regex when one doesn't have to.

Show quoteHide quote

"lgbjr" wrote:

> Dennis,
>
> Previously, that is exactly what I was doing:
>
> Dim SA as Array
> Dim S as String
> S="FBE"
> SA=S.ToCharArray
>
> This works fine. but, I'm trying to use Options Strict now, which means I
> can't use
>
> Dim SA as Array
>
> I have to use
>
> Dim SA as String()
>
> And 1-dimensional array of Char can not be converted to 1-dimensional array
> of String. So, I decided to use a Regex Match to convert the string to a
> string array.
>
> LOL! As I was typing this, I just realized I can do Dim SA as Char(), then
> use .ToCharArray!!
>
> Thanks!!
>
> Lee
>
> "Dennis" <Den***@discussions.microsoft.com> wrote in message
> news:1D85F3C7-FC03-4B75-BA56-109445A9BF92@microsoft.com...
> > Why can't you use the .ToCharArray method of Strings?
> >
> > "lgbjr" wrote:
> >
> >> Thanks Jay,
> >>
> >> I'm an idiot!! an hour or so ago, I was working on a split (really a
> >> split),
> >> and just continued with split when I should have been using Matches. LOL!
> >> Of
> >> course, Expresso was giving me the result I expected, because it wasn't
> >> trying to split the string!
> >>
> >> thanks for pointing out what should have been an obvious mistake.
> >>
> >> Lee
> >>
> >> "Jay B. Harlow [MVP - Outlook]" <Jay_Harlow_***@msn.com> wrote in message
> >> news:%23JuWn9CQFHA.3144@tk2msftngp13.phx.gbl...
> >> > Lee,
> >> > | I expect that SA will contain 3 elements: SA(0)="F", SA(1)="B",
> >> > SA(2)="E",
> >> > | but, what I'm getting is 7 elements: SA(0)="", SA(1)="F", SA(2)="",
> >> > | SA(3)="B", SA(4)="", SA(5)="E", SA(6)="".
> >> > I would expect it to contain 4 elements, SA(0)="", SA(1)="", SA(2)="",
> >> > and
> >> > SA(3)="", as your string only contains delimiters. RegEx.Split returns
> >> > the
> >> > strings between the delimiters, unless you use capturing groups (the
> >> > parenthesis in your expression) in which case it returns both the
> >> > strings
> >> > between the delimiters & the delimiters.
> >> >
> >> > The pattern "[A-F,0-9]" returns the 4 that I expect. The capturing
> >> > groups
> >> > in
> >> > your expression is causing RegEx to return the 4 strings between the
> >> > delimiters, plus the 3 delimiters, ergo 7 values.
> >> >
> >> >
> >> > It sounds like you really want to return the list of matches, rather
> >> > then
> >> > the stuff between the delimiters... Try RegEx.Matches, something like:
> >> >
> >> >        Dim input As String = "FBE"
> >> >
> >> >        Const pattern As String = "([A-F,0-9])"
> >> >        Static parser As New Regex(pattern)
> >> >
> >> >        For Each match As Match In parser.Matches(input)
> >> >            Debug.WriteLine(match.Value)
> >> >        Next
> >> >
> >> > Hope this helps
> >> > Jay
> >> >
> >> >
> >> >
> >> > "lgbjr" <lgbjr@online.nospam> wrote in message
> >> > news:OlvF7wCQFHA.2356@TK2MSFTNGP14.phx.gbl...
> >> > | Hi All,
> >> > |
> >> > | I'm trying to split a string on every character. The string happens
> >> > to
> >> > be
> >> > a
> >> > | representation of a hex number. So, my regex expression is
> >> > ([A-F,0-9]).
> >> > | Seems simple, but for some reason, I'm not getting the results I
> >> > expect.
> >> > |
> >> > | Dim SA as string()
> >> > | Dim S as string
> >> > |
> >> > | S="FBE"
> >> > | SA=RegularExpressions.Regex.Split(S,"([A-F,0-9])")
> >> > |
> >> > | I expect that SA will contain 3 elements: SA(0)="F", SA(1)="B",
> >> > SA(2)="E",
> >> > | but, what I'm getting is 7 elements: SA(0)="", SA(1)="F", SA(2)="",
> >> > | SA(3)="B", SA(4)="", SA(5)="E", SA(6)="".
> >> > |
> >> > | If I change the expression to [A-F,0-9] (no parentheses), I get:
> >> > SA(0)="",
> >> > | SA(1)="", SA(2)="", SA(3)="".
> >> > |
> >> > | Just for my own sanity, I've checked the pattern in Expresso and it
> >> > returns
> >> > | what I would expect.
> >> > |
> >> > | I suppose I should mention I'm using VB.NET 2005 (just in case
> >> > there's a
> >> > | known issue with Regex in 2005).
> >> > |
> >> > | TIA
> >> > | Lee
> >> > |
> >> > |
> >> >
> >> >
> >>
> >>
> >>
>
>
>