vim search and replace between number

Question

I have a pattern where there are double-quotes between numbers in a CSV file. I can search for the pattern by [0-9]\"[0-9], but how do I retain value while removing the double quote. CSV format is like this:

"1234"5678","Text1","Text2"
"987654321","Text3","text4"
"7812891"3","Text5","Text6"

As you may notice there are double quotes between some numbers which I want to remove.

I have tried the following way, which is incorrect:

:%s/[0-9]\"[0-9]/[0-9][0-9]/g

Is it possible to execute a command at every search pattern, maybe go one character forward and delete it. How can "lx" be embedded in search and replace.

Answer 1

You need to capture groups. Try:

:%s/\(\d\)"\(\d\)/\1\2/g

[A digit can also be denoted by \d.]

Answer 2

I know that this question has been answered already, but here's another approach:

:%s/\d\zs"\ze\d

Explanation:

%s   Substitute for the whole buffer
\d   look up for a digit
\zs set the start of match here
"     look up for a double-quote
\ze set the end of match here
\d   look up for a digit

That makes the substitute command to match only the double-quote surrounded by digits.
Omitting the replacement string just deletes the match.