Regular Expression - Match all except last one

[Randy Webb]

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 = "asdfasdfasdfasdf";
var newString = myString.replace(pattern,'gh');
alert(newString)

Gives me: asghasghasghasgh as it should.
What I want: asghasghasghasdf Where the last one is not replaced.

 

[BootNic]

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 = "asdfasdfasdfasdf";
var newString = myString.replace(pattern,'gh');
alert(newString)

Gives me: asghasghasghasgh as it should.
What I want: asghasghasghasdf 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*

 

[Randy Webb]

BootNic said the following on 3/19/2006 1:20 AM:

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.

 

[BootNic]

BootNic said the following on 3/19/2006 1:20 AM:

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 = "asdfasdfasdfasdf";
var newString = myString.replace(pattern,'gh');
alert(newString);

Or

var myString = "asdfasdfasdfasdf";
while(myString.match(/df/g).length!=1){
myString=myString.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 *

 

[Alexander Bartolich]

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 = "asdfasdfasdfasdf";
var newString = myString.replace(pattern,'gh');
alert(newString)

Gives me: asghasghasghasgh as it should.
What I want: asghasghasghasdf Where the last one is not replaced.

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

--

 

[Randy Webb]

Alexander Bartolich said the following on 3/19/2006 4:33 PM:

var newString = myString.replace(/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.

 

[Dr John Stockton]

JRS: In article <[email protected]>, dated Sun, 19 Mar
2006 02:04:22 remote, seen in Randy Webb

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.

 

[Alexander Bartolich]

Alexander Bartolich said the following on 3/19/2006 4:33 PM:

[...]
var newString = myString.replace(/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.

 

[Randy Webb]

Alexander Bartolich said the following on 3/20/2006 3:03 AM:

Alexander Bartolich said the following on 3/19/2006 4:33 PM:

[...]
var newString = myString.replace(/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 Webb]

Randy Webb said the following on 3/20/2006 6:58 AM:

Alexander Bartolich said the following on 3/20/2006 3:03 AM:

Alexander Bartolich said the following on 3/19/2006 4:33 PM:
[...]
var newString = myString.replace(/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.

I think I spoke too soon for it doing exactly what I want it to do. I
think I gave a bad example with my first question as it doesn't actually
do what I need it to do in the real scenario.

This is my actual code:

var input = document.myForm.textarea1;
var output = document.myForm.textarea2;
delim = /\r\n(.)/g;
output.value = "document.write('" +
input.value.replace(delim,'\');\ndocument.write\(\'$1');
output.value = output.value + "');";

What it does is take the input value of one textarea and create
document.write statements with every line. It is fine unless there is an
extra line at the end.

If the value of textarea1 is:

test<nothing more>

Then it works properly and give this in textarea2:

document.write('test');

But, if textarea1 looks like this:

test
<nothing more>

Where it has that extra carriage return then the code I get is this:

document.write('test
');

Which isn't what I want.

So, how do I deal with that extra line in the textarea?

 

[BootNic]

news:[email protected].... [snip]
So, how do I deal with that extra line in the textarea?

Trim the white space.

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"
"http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<script type="text/javascript">
function tellerror(msg, url, lnn){
alert('Error message= '+msg+'\nURL= '+url+'\nLine Number= '+lnn)
return true
}
window.onerror=tellerror
</script>
<script type="text/javascript">
function cod(){
String.prototype.trim=function(){return this.replace(/^\s+|\s+$/g,'')};
var input = document.myForm.textarea1;
var output = document.myForm.textarea2;
delim = /(\r\n)+|(\n)+/g;
output.value='';
output.value="document.write('" +
input.value.trim().replace(delim,'\');\ndocument.write\(\'')+ "');"
}
</script>
<title></title>
</head>
<body>
<form action="javascript:cod()" name="myForm" id="myForm">
<textarea wrap="off" name="textarea1" cols="25" rows="15">
First line.






Second Line.
Third Line.


Last Line.



</textarea>
<textarea wrap="off" name="textarea2" cols="25" rows="15">
</textarea> <input type="submit" value="cod">
</form>
</body>
</html>


--
BootNic Monday, March 20, 2006 8:24 AM

Sometimes I think the surest sign that intelligent life exists
elsewhere in the universe is that none of it has tried to contact us.
*Bill Watterson*

 

[Richard Lionheart]

Hi BootNic,

Great code enhancement. I'm an html/ javascript noobie, so I like
trolling the Q&As on this newsgroup to learn coding techniques.

Question: Do you concur in my belief that the preliminary null
assignment to output.value is superfluous because the succeeding
assignment completely replaces output.value's content?

Question: What's a good on-line source for learning the semantics of
your "String.prototype.trim" assignment statement? I don't mean the RE
stuff or replacing stings of leading or trailing space characters with
nulls. I mean:
- Does String have a property that has a trim method;
- Or is prototype a generic method to modify any method of any class

Thanks in advance,
Richard

 

[Lasse Reichstein Nielsen]

Alexander Bartolich said the following on 3/19/2006 4:33 PM:

[...]
var newString = myString.replace(/df(.)/g,'gh$1');

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

And it will fail on "dfdfdf", replacing only the first one.

If you expect the regexp engine to be recent, you can use positive
lookahead:
/df(?:.)/g
Then you could also just use negative lookahead on the end of text:
/df(?!$)/g

It's quite a lot harder without lookahead, since regular expressions
are notoriously hard to get to do a negative match (which is what
"not at the end" is).

/L

 

[BootNic]

Hi BootNic,

Great code enhancement. I'm an html/ javascript noobie, so I like
trolling the Q&As on this newsgroup to learn coding techniques.

Question: Do you concur in my belief that the preliminary null
assignment to output.value is superfluous because the succeeding
assignment completely replaces output.value's content?

It should not be there, I would guess that I started to type out something
different then desided to paste instead.

Question: What's a good on-line source for learning the semantics of
your "String.prototype.trim" assignment statement? I don't mean
the RE stuff or replacing stings of leading or trailing space
characters with nulls. I mean:
- Does String have a property that has a trim method;
- Or is prototype a generic method to modify any method of any class

I would not normally use prototype, I used it this time simply because there
is something similar in the FAQ. It also should be outside of the function.

I prefer the Trim(str) function over the prototype. In example I posted,
a replace(/^\s+|\s$/g,'') in place of trim() would have done well.

http://jibbering.com/faq/#FAQ4_16

There is a lot of room for improvement on what I posted.

for example delim = /(\r\n)+|(\n)+/g; could be written different.

--
BootNic Monday, March 20, 2006 3:00 PM

"This is all very interesting, and I daresay you already see me
frothing at the mouth in a fit; but no, I am not; I am just winking
happy thoughts into a little tiddle cup."
*Nabokov, Lolita*

 

[Dr John Stockton]

JRS: In article <[email protected]>, dated Mon, 20 Mar
2006 07:05:16 remote, seen in Randy Webb

So, how do I deal with that extra line in the textarea?

You could, maybe, .split the textarea at newlines; then scan the array
dealing with empty lines (e.g. put only non-empty ones into another
array, or remove trailing empties), also adding the ...(" and ");
strings to each element, then rejoin. It's not what you asked for, but
it might be clearer.

<URL:http://www.merlyn.demon.co.uk/js-quick.htm> Pack uses this

function Pak(F) { var S = F.Code.value, L = +F.Len.value, J, Re
S = S.replace(/(\r?\n){2,}/g, "\u2029") // PS
S = S.replace(/(\r?\n)/g, " ")
S = S.replace(/\s+/g, " ").split("\u2029")
if (L<1) L = F.Len.value = 1
Re = new RegExp("(.{1," + L + "}) ", "g")
J = S.length ; while (J--)
S[J] = (S[J].replace(/^\s+/, "")+" ").replace(Re, "$1\n")
F.Code.value = S.join("\n") }

to paragraph-pack textarea Code; but I didn't care about non-visible
characters in the output. You might get an idea from it. Or suggest an
improvement.

 

[Randy Webb]

Dr John Stockton said the following on 3/20/2006 6:20 PM:

JRS: In article <[email protected]>, dated Mon, 20 Mar
2006 07:05:16 remote, seen in Randy Webb


You could, maybe, .split the textarea at newlines; then scan the array
dealing with empty lines (e.g. put only non-empty ones into another
array, or remove trailing empties), also adding the ...(" and ");
strings to each element, then rejoin. It's not what you asked for, but
it might be clearer.

Ironically enough, that is exactly how I wrote it to start with and a
thread a few days ago about ABCEF: or something like that made me
start wondering if I could hack together a RegEx solution so I started
whacking. In the end, I went back to exactly what you suggested (that I
started out with).

Uncommented, this is the entire function:

function convertIt(){
var output = document.myForm.textarea2;
var myArray =
document.myForm.textarea1.value.replace('\'','\\\'').split('\r\n');
for (var i=0;i<myArray.length;i++)
if (myArray != '')
{
output.value += "document.write('" + myArray + "');\r\n";
}
}

There are a few more things I am going to add to it, such as removing
double blank spaces, and empty lines with just a space in it. But
compared to the RegExp solutions so far, the above is - by far - simpler
and clearer (as you suggested).

<URL:http://www.merlyn.demon.co.uk/js-quick.htm> Pack uses this

function Pak(F) { var S = F.Code.value, L = +F.Len.value, J, Re
S = S.replace(/(\r?\n){2,}/g, "\u2029") // PS
S = S.replace(/(\r?\n)/g, " ")
S = S.replace(/\s+/g, " ").split("\u2029")
if (L<1) L = F.Len.value = 1
Re = new RegExp("(.{1," + L + "}) ", "g")
J = S.length ; while (J--)
S[J] = (S[J].replace(/^\s+/, "")+" ").replace(Re, "$1\n")
F.Code.value = S.join("\n") }

to paragraph-pack textarea Code; but I didn't care about non-visible
characters in the output. You might get an idea from it. Or suggest an
improvement.

I have saved it for future reference but admittedly, my knowledge of
RegExp's is not such that I could comment on it, mostly ask questions.

 

[Dr John Stockton]

JRS: In article <[email protected]>, dated Tue, 21 Mar
2006 01:15:47 remote, seen in Randy Webb
PS is IIRC Paragraph-Space, represented by \u2029. Multiple blank lines
are turned into PS.
Remaining newlines are turned into spaces.
Whitespace is turned into single-space, and each paragraph becomes an
array element.
The minimum line length is one character.
Re matches any number of characters from 1 to L followed by a space, and
remembers the characters.
For each paragraph

S[J] = (S[J].replace(/^\s+/, "")+" ").replace(Re, "$1\n")

remove leading spaces, add one trailing space[*]; replace, with a
newline, each occurrence of a space preceded with enough characters[+].

Join the paragraphs.

I have saved it for future reference but admittedly, my knowledge of
RegExp's is not such that I could comment on it, mostly ask questions.

[*] H'mmm - in principle, maybe I should be removing initial trailing
spaces here. It appears unnecessary.
[+] That code is right; but I'm not entirely sure why.

The code of the page is now better; I've added, first,
S = S.replace(/^\s+|\s+$/g, "") // FAQ 4.15 Trim
to clean the ends.