Vista - regular expressions to replace but keep character?

07-15-2008

Is there a way to use Replace with Regular Expressions but keep a character
in the original string? A simplified example: If we have "< a >" or ""
or " " within a string and we want to remove the spaces, but keep the
character, is there a way this can be done? A Regular Expression can find
the substring with something like "<\s*[a-z]\s*>" (only finds one character),
but the ReplaceWith argument appears to be static. Is there any way to keep
the character?

Thanks!,
Nick

07-15-2008

Nick B schrieb:

Quote:

> Is there a way to use Replace with Regular Expressions but keep a character
> in the original string? A simplified example: If we have "< a >" or ""
> or " " within a string and we want to remove the spaces, but keep the
> character, is there a way this can be done? A Regular Expression can find
> the substring with something like "<\s*[a-z]\s*>" (only finds one character),
> but the ReplaceWith argument appears to be static. Is there any way to keep
> the character?

[...]
You can use $<n> (<n> = number of () in pattern starting with 1) in the
replacement string:

oRE.Pattern = "(<)(\s*)(\S+)(\s*)(>)"
1 2 3 4 5
sRes = oRE.Replace( " ", "$1$3$5" )

07-15-2008

That's exactly what I couldn't find. Thanks!!

"ekkehard.horner" wrote:

Quote:

> Nick B schrieb:

Quote:

> > Is there a way to use Replace with Regular Expressions but keep a character
> > in the original string? A simplified example: If we have "< a >" or ""
> > or " " within a string and we want to remove the spaces, but keep the
> > character, is there a way this can be done? A Regular Expression can find
> > the substring with something like "<\s*[a-z]\s*>" (only finds one character),
> > but the ReplaceWith argument appears to be static. Is there any way to keep
> > the character?

> [...]
> You can use $<n> (<n> = number of () in pattern starting with 1) in the
> replacement string:
>
> oRE.Pattern = "(<)(\s*)(\S+)(\s*)(>)"
> 1 2 3 4 5
> sRes = oRE.Replace( " ", "$1$3$5" )
>

07-15-2008

"ekkehard.horner" wrote:

Quote:

>
> oRE.Pattern = "(<)(\s*)(\S+)(\s*)(>)"
> 1 2 3 4 5
> sRes = oRE.Replace( " ", "$1$3$5" )
>

You can do it a lot simpler than that.

oRE.Pattern = "<\s*([^\s>]+)\s*>"
sRes = oRE.Replace( " ", "<$1>" )

Since you *KNOW* that you are looking for <...> and you *KNOW* you want to
retain <...>, just do *NOT* put any of that stuff in the parentheses. So now
only the non-space, non-> characters will be in parens and there's only one
"group" to be plunked into the output.

07-15-2008	#1 (permalink)
Nick B n/a posts	regular expressions to replace but keep character? Is there a way to use Replace with Regular Expressions but keep a character in the original string? A simplified example: If we have "< a >" or "<b >" or "< br>" within a string and we want to remove the spaces, but keep the character, is there a way this can be done? A Regular Expression can find the substring with something like "<\s[a-z]\s>" (only finds one character), but the ReplaceWith argument appears to be static. Is there any way to keep the character? Thanks!, Nick
My System Specs

07-15-2008	#2 (permalink)
ekkehard.horner n/a posts	Re: regular expressions to replace but keep character? Nick B schrieb: Quote: > Is there a way to use Replace with Regular Expressions but keep a character > in the original string? A simplified example: If we have "< a >" or "<b >" > or "< br>" within a string and we want to remove the spaces, but keep the > character, is there a way this can be done? A Regular Expression can find > the substring with something like "<\s[a-z]\s>" (only finds one character), > but the ReplaceWith argument appears to be static. Is there any way to keep > the character? [...] You can use $<n> (<n> = number of () in pattern starting with 1) in the replacement string: oRE.Pattern = "(<)(\s)(\S+)(\s)(>)" 1 2 3 4 5 sRes = oRE.Replace( "< br >", "$1$3$5" )
My System Specs

07-15-2008	#3 (permalink)
Nick B n/a posts	Re: regular expressions to replace but keep character? That's exactly what I couldn't find. Thanks!! "ekkehard.horner" wrote: Quote: > Nick B schrieb: Quote: > > Is there a way to use Replace with Regular Expressions but keep a character > > in the original string? A simplified example: If we have "< a >" or "<b >" > > or "< br>" within a string and we want to remove the spaces, but keep the > > character, is there a way this can be done? A Regular Expression can find > > the substring with something like "<\s[a-z]\s>" (only finds one character), > > but the ReplaceWith argument appears to be static. Is there any way to keep > > the character? > [...] > You can use $<n> (<n> = number of () in pattern starting with 1) in the > replacement string: > > oRE.Pattern = "(<)(\s)(\S+)(\s)(>)" > 1 2 3 4 5 > sRes = oRE.Replace( "< br >", "$1$3$5" ) >
My System Specs

07-15-2008	#4 (permalink)
Old Pedant n/a posts	Re: regular expressions to replace but keep character? "ekkehard.horner" wrote: Quote: > > oRE.Pattern = "(<)(\s)(\S+)(\s)(>)" > 1 2 3 4 5 > sRes = oRE.Replace( "< br >", "$1$3$5" ) > You can do it a lot simpler than that. oRE.Pattern = "<\s([^\s>]+)\s>" sRes = oRE.Replace( "< br >", "<$1>" ) Since you KNOW that you are looking for <...> and you KNOW you want to retain <...>, just do NOT put any of that stuff in the parentheses. So now only the non-space, non-> characters will be in parens and there's only one "group" to be plunked into the output.
My System Specs

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