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Regular Expression - Match all except last one

I know that the /g flag will match all occurrences. Is there a way, with
a Regular Expression, to match all occurrences *except* the last one?

pattern = /df/g;
var myString = "asdfasdfasdfas df";
var newString = myString.replac e(pattern,'gh') ;
alert(newString )

Gives me: asghasghasghasg h as it should.
What I want: asghasghasghasd f Where the last one is not replaced.

--
Randy
comp.lang.javas cript FAQ - http://jibbering.com/faq & newsgroup weekly
Javascript Best Practices - http://www.JavascriptToolbox.com/bestpractices/
Mar 19 '06 #1
17 11659
> "Randy Webb" <Hi************ @aol.com> wrote:
news:Q7******** ************@co mcast.com....

I know that the /g flag will match all occurrences. Is there a way,
with a Regular Expression, to match all occurrences *except* the
last one?

pattern = /df/g;
var myString = "asdfasdfasdfas df";
var newString = myString.replac e(pattern,'gh') ;
alert(newString )

Gives me: asghasghasghasg h as it should.
What I want: asghasghasghasd f Where the last one is not replaced.


Does the string always end with the pattern?

--
BootNic Sunday, March 19, 2006 1:19 AM

It's not that some people have willpower and some don't. It's that
some people are ready to change and others are not.
*James Gordon*

Mar 19 '06 #2
BootNic said the following on 3/19/2006 1:20 AM:
"Randy Webb" <Hi************ @aol.com> wrote:
news:Q7******** ************@co mcast.com....

I know that the /g flag will match all occurrences. Is there a way,
with a Regular Expression, to match all occurrences *except* the
last one?

pattern = /df/g;
var myString = "asdfasdfasdfas df";
var newString = myString.replac e(pattern,'gh') ;
alert(newString )

Gives me: asghasghasghasg h as it should.
What I want: asghasghasghasd f Where the last one is not replaced.


Does the string always end with the pattern?


Yes. And I think I know where you are headed but I will wait and see :)

The pattern I am actually matching is \r\n and the string is the .value
of a textarea. It is replacing it with ";\r\ndocument. write("
Where the " is part of the replacement. All it does is take code and
create document.write statements for them. What I end up with at the end
is an extra document.write( " that I don't want.

I could always just replace the last occurence of document.write( " but I
was hoping there was a simpler solution.
--
Randy
comp.lang.javas cript FAQ - http://jibbering.com/faq & newsgroup weekly
Javascript Best Practices - http://www.JavascriptToolbox.com/bestpractices/
Mar 19 '06 #3
> "Randy Webb" <Hi************ @aol.com> wrote:
news:qZ******** ************@co mcast.com....

BootNic said the following on 3/19/2006 1:20 AM:
"Randy Webb" <Hi************ @aol.com> wrote:
news:Q7******** ************@co mcast.com....

I know that the /g flag will match all occurrences. Is there a
way, with a Regular Expression, to match all occurrences *except*
the last one?

pattern = /df/g;
var myString = "asdfasdfasdfas df";
var newString = myString.replac e(pattern,'gh') ;
alert(newString )

Gives me: asghasghasghasg h as it should.
What I want: asghasghasghasd f Where the last one is not replaced.


Does the string always end with the pattern?


Yes. And I think I know where you are headed but I will wait and
see :)

The pattern I am actually matching is \r\n and the string is the
.value of a textarea. It is replacing it with ";\r\ndocument. write("
Where the " is part of the replacement. All it does is take code and
create document.write statements for them. What I end up with at
the end is an extra document.write( " that I don't want.

I could always just replace the last occurence of document.write( "
but I was hoping there was a simpler solution.


Well now, I don't think I am comprehending the situation. It sounds like
you need to strip the the white space on each end of the sting before the
replace.

But I don't think that is the issue.

I was going to suggest one of two things.

pattern = /df(?=.)/g;
var myString = "asdfasdfasdfas df";
var newString = myString.replac e(pattern,'gh') ;
alert(newString );

Or

var myString = "asdfasdfasdfas df";
while(myString. match(/df/g).length!=1){
myString=myStri ng.replace(/df/,'gh')
}
alert(myString) ;

--
BootNic Sunday, March 19, 2006 2:21 AM

Get your facts first, and then you can distort them as much as you
please.
*Mark Twain *

Mar 19 '06 #4
Randy Webb wrote:
I know that the /g flag will match all occurrences. Is there a way, with
a Regular Expression, to match all occurrences *except* the last one?

pattern = /df/g;
var myString = "asdfasdfasdfas df";
var newString = myString.replac e(pattern,'gh') ;
alert(newString )

Gives me: asghasghasghasg h as it should.
What I want: asghasghasghasd f Where the last one is not replaced.


var newString = myString.replac e(/df(.)/g,'gh$1');

--
Mar 19 '06 #5
Alexander Bartolich said the following on 3/19/2006 4:33 PM:
Randy Webb wrote:
I know that the /g flag will match all occurrences. Is there a way, with
a Regular Expression, to match all occurrences *except* the last one?

pattern = /df/g;
var myString = "asdfasdfasdfas df";
var newString = myString.replac e(pattern,'gh') ;
alert(newString )

Gives me: asghasghasghasg h as it should.
What I want: asghasghasghasd f Where the last one is not replaced.


var newString = myString.replac e(/df(.)/g,'gh$1');


What do the (.) and $1 do? The code does exactly what I want it to do
but I want to understand it now.

--
Randy
comp.lang.javas cript FAQ - http://jibbering.com/faq & newsgroup weekly
Javascript Best Practices - http://www.JavascriptToolbox.com/bestpractices/
Mar 19 '06 #6
JRS: In article <qZ************ ********@comcas t.com>, dated Sun, 19 Mar
2006 02:04:22 remote, seen in news:comp.lang. javascript, Randy Webb
<Hi************ @aol.com> posted :
The pattern I am actually matching is \r\n and the string is the .value
of a textarea. It is replacing it with ";\r\ndocument. write("
Where the " is part of the replacement. All it does is take code and
create document.write statements for them. What I end up with at the end
is an extra document.write( " that I don't want.

I could always just replace the last occurence of document.write( " but I
was hoping there was a simpler solution.


Append the string '") ;' and you have effectively neutralised it.

Or can you pre-process myString with a .replace that removes just the
final \r\n? Or chop off the last 2 characters with substr or substring?
OTOH, can you not replace \r\n(.) with ";\r\ndocument. write("$1
which should not see a final \r\n ? Untested.

--
© John Stockton, Surrey, UK. ?@merlyn.demon. co.uk Turnpike v4.00 IE 4 ©
<URL:http://www.jibbering.c om/faq/> JL/RC: FAQ of news:comp.lang. javascript
<URL:http://www.merlyn.demo n.co.uk/js-index.htm> jscr maths, dates, sources.
<URL:http://www.merlyn.demo n.co.uk/> TP/BP/Delphi/jscr/&c, FAQ items, links.
Mar 19 '06 #7
Dr John Stockton said the following on 3/19/2006 6:35 PM:
JRS: In article <qZ************ ********@comcas t.com>, dated Sun, 19 Mar
2006 02:04:22 remote, seen in news:comp.lang. javascript, Randy Webb
<Hi************ @aol.com> posted :
The pattern I am actually matching is \r\n and the string is the .value
of a textarea. It is replacing it with ";\r\ndocument. write("
Where the " is part of the replacement. All it does is take code and
create document.write statements for them. What I end up with at the end
is an extra document.write( " that I don't want.

I could always just replace the last occurence of document.write( " but I
was hoping there was a simpler solution.
Append the string '") ;' and you have effectively neutralised it.


That is, to date, the best suggestion I have seen/read :) I already
prepend document.write( ' to it so appending isn't a big deal.
Or can you pre-process myString with a .replace that removes just the
final \r\n? Or chop off the last 2 characters with substr or substring?
That was also something I considered. Check to see if the last two are
\r\n (is that 2 or 4 characters?) and go from there. I need to do
different things if there is a final \r\n with no text following it.

Something else I considered was adding a final \r\n and then remove any
empty document.write statements (just so they aren't there, they won't
hurt anything).
OTOH, can you not replace \r\n(.) with ";\r\ndocument. write("$1
which should not see a final \r\n ? Untested.


It doesn't work correctly if there is a final \r\n in the string.

delim = /\r\n(.)/g;
output.value="d ocument.write(' "+input.value.r eplace(delim,'\ ');\ndocument.w rite\(\'$1');

output and input are references to a textarea.
If the final line of the input is an empty line, where you type in text
and then press the Enter key with no more text, it breaks horribly. As
long as there is no \r\n at the very end of the string, it works perfectly.
--
Randy
comp.lang.javas cript FAQ - http://jibbering.com/faq & newsgroup weekly
Javascript Best Practices - http://www.JavascriptToolbox.com/bestpractices/

Mar 20 '06 #8
Randy Webb wrote:
Alexander Bartolich said the following on 3/19/2006 4:33 PM:
[...]
var newString = myString.replac e(/df(.)/g,'gh$1');


What do the (.) and $1 do? The code does exactly what I want
it to do but I want to understand it now.


The dot matches any character.
The round brackets form a parenthesized subexpression.
The text matched by such subexpression is available in the
replacement text as $1, $2, $3, etc.

So what this does is to search for "df" followed by any character,
and replace that with "gh", followed by that character.

--
post tenebras lux. post fenestras tux.
Mar 20 '06 #9
Alexander Bartolich said the following on 3/20/2006 3:03 AM:
Randy Webb wrote:
Alexander Bartolich said the following on 3/19/2006 4:33 PM:
[...]
var newString = myString.replac e(/df(.)/g,'gh$1');

What do the (.) and $1 do? The code does exactly what I want
it to do but I want to understand it now.


The dot matches any character.
The round brackets form a parenthesized subexpression.
The text matched by such subexpression is available in the
replacement text as $1, $2, $3, etc.

So what this does is to search for "df" followed by any character,
and replace that with "gh", followed by that character.


Thank you for the code and the explanation. I guess it's finally time
for me to read my references on Regular Expressions, asking questions,
and learn the things.

--
Randy
comp.lang.javas cript FAQ - http://jibbering.com/faq & newsgroup weekly
Javascript Best Practices - http://www.JavascriptToolbox.com/bestpractices/
Mar 20 '06 #10

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