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How can I use regular expressions to replace the text [Backspace] with a real backspace? I can find it with \[Backspace\] and have tried to replace it with [\b] and \b with POSIX both on and off. Each time it simply replaces it literally with [\b] or \b.

I think you may actually need to replace the preceeding character with nothing. Something like this:

Search for: .{1}\[backspace\]
Replace with: nothing

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Testing:
Starting with this:
abcdefgp[backspace]hijklm

Results with this:
abcdefghijklm

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Using Regular Expression with POSIX

This example will have problems if multiple backspaces like this:
abcdefwp[backspace][backspace]ghijklm

To insert a literal backspace use the replacement expression \x08.

I do have multiples as you suggested. It fails on those since the cursor position is already past the previous character. Therefore, after the second iteration of the replace on [Backspace][Backspace][Backspace], it looks like this [Backspace since it selects [Backspace][Backspace].

It somehow needs to replace it with a backspace so that the cursor position does not bypass it. I also need it to do the real backspace so that it removes the previous character as well... as you were trying to do.

ben, That replaces it with a block symbol but doesn't actually perform the backspace. Using Bob's suggestion on this then gets me the same results.

Is there a way to find any number of occurences of something, then grab the same number or prior characters, then replace them all with nothing?

Find Replace with nothing
Repeat until the replace operation gives an error "not found".

This transforms

xxxabc[Backspace][Backspace][Backspace]def

in the first step to

xxxab[Backspace][Backspace]def

in the second step to

xxxa[Backspace]def

and after the third step

xxxdef

As regexes can't count, I see no singlestep solution.

(as always, POSIX syntax is used)

The block symbol represents a "real backspace" (0x08), which is what you orginally asked for.

What you are looking for can't be done with a single application of a regular expression, as regular expression recognisers "can't count". But once you have replaced [backspace] with any single character of your choice (say # (where # doesn't occur elsewhere in the file)), you can repeatedly replace [^#]# by nothing.

The advantage of initially replacing [backspace] with a single character is that the subsequent repeated replacement will then correctly handle a backspace following a "genuine" ].

Ben, I realize that the block symbol was the unprintable backspace character, but I was hoping it could actually do the backspace. MudGuard's (and your) suggestion to use [^...] instead of .{1} works. I simply keep hitting "Replace All" until it can't find any more. Thanks to all

You could also handle multiple backspaces by making multiple passes with a modifed version of my first suggestion with one for three backspaces, then two backspaces, and then one backspace.

Untested:
Search for: .{3}(\[backspace\]){3}
Replace with: nothing

Search for: .{2}(\[backspace\]){2}
Replace with: nothing

Search for: .{1}\[backspace\]
Replace with: nothing

If this works, then make a macro to run them in sequence.

Bob, in your case, the maximum number of consecutive backspaces is limited

e.g. if you only use 3, but there are seven, this:
987654321[backspace][backspace][backspace][backspace][backspa0ce][backspace][backspace]
produces
987654[backspa
when doing the 3-backspaces deleting

More general: if the longest sequence you program for is n but there is a sequence of 2n+1, your method fails.