Regex to find double whitespace not on the start of a line

Question

I'm struggling with old "formatted" code, where a lot of whitespace is added to line up ='s and so on:

  if (!This.RevData.Last().Size() eq 0) then
    !DocRev   = '?'
    !Status   = '?'
    !RevCode  = '?'
  else
    !Pad      = !This.RevData.Last()[1]
    !DocRev   = !Pad[2]
    !Status   = !This.GenUtil.If((!Pad[3] eq 'UA'), '' , !Pad[3])
    !RevCode  = !This.GenUtil.If((!Pad[6] eq ''  ), '?', !Pad[6])
  endif

In this example it actually makes some sense, but most often the code have been modified to a state that makes the whitespace much more confusing than helpful.

What I'd like to do is to replace all double (or more) whitespaces with a single space, but I'd of course like to keep the indenting. Hence I'm looking for a regex to identify double (or more) spaces that are not at the start of the line. I've tried negative lookbehind, but can't seem to make it work:

(?<![\n])[\s]{2,}

Any hints, anyone? Ohh, and I'm using Ultraedit (the Perl regex engine), so the regex should be "UE-compatible".

EDIT: UE doesn't evaluate regex's line for line; newlines is just a character in the long string that is the document, which complicates the problem a bit.

Answer 1

Replace "([^ \n]) +" with "$1". No funky lookbehinds required.

(For emphasis, the markup doesn't show it clearly, but there are two spaces before the plus sign, to avoid needlessly replacing single spaces with single spaces.)

Answer 2

It's not PCRE but this will do what you need if you have access to a Linux shell:

sed s/"([^ ]) +"/"\1 "/g source.code > reformatted.code

It will just replace any spaces that follow a non-space character while preserving that character. Should be easy enough to perlify it, if you're used to Perl Regexs.

Answer 3

Try replacing...

(?<=[^\r\n])([\t ])[\t ]*

with...

$1