Regular expression to find and replace unescaped Non-successive double quotes in CSV file

Question

This is an extension to a related question answered Here

I have a weekly csv file which needs to be parsed. it looks like this.

\"asdf\",\"asdf\",\"as

Answer 1

I was using VIM to remove nested quotes in a .CSV file and this worked for me:

"[^,"][^"]*"[^,]

Answer 2

(?<!^|,)"(?!,|$)

will match a double quote that is not preceded or followed by a comma nor situated at start/end of line.

If you need to allow whitespace around the commas or at start/end-of-line, and if your regex flavor (which you didn't specify) allows arbitrary-length lookbehind (.NET does, for example), you can use

(?<!^\s*|,\s*)"(?!\s*,|\s*$)

Answer 3

In vim I used this to remove all the unescaped quotes.

:%s/\v("(,")@!)&((",)@<!")&("(\n)@!)&(^@<!")//gc

detailed explanation is,

: - start the vim command
    % - scope of the command is the whole file
    s - search and replace
        / - start of search pattern
        \v - simple regex syntax (rather than vim style)
            (
                " - double quote
                (,") - comma_quote
                @! - not followed by
            )
            & - and
            (
                (",) - quote_comma
                @<!- does not precedes
                " - double quote
            )
            & - and
            (
                " - double quote
                (\n) - line end
                @! - not followed by
            )
            & - and
            (
                ^ - line beginning
                @<! - does not precedes
                " - double quote
            )
        / - end of search pattern and start of replace pattern
             - replace with nothing (delete)
        / - end of replace pattern
    g - apply to all the matches
    c - confirm with user for every replacement

this does the job fairly quickly. The only instance this fails is when there are instances of "," pattern in the data.