Modify content inside quotation marks, BASH

Question

Good day to all,

I was wondering how to modify the content inside quotation marks and left unmodified the outside.

Input line:

,,,"Investigacion,,, desarrollo",,,

Output line:

,,,"Investigacion, desarrollo",,,

Initial try:

sed 's/\"",,,""*/,/g' 

But nothing happens, thanks in advance for any clue

Answer 1

The idiomatic awk way to do this is simply:

$ awk 'BEGIN{FS=OFS="\""} {sub(/,+/,",",$2)} 1' file
,,,"Investigacion, desarrollo",,,

or if you can have more than one set of quoted strings on each line:

$ cat file
,,,"Investigacion,,, desarrollo",,,"foo,,,,bar",,,

$ awk 'BEGIN{FS=OFS="\""} {for (i=2;i<=NF;i+=2) sub(/,+/,",",$i)} 1' file
,,,"Investigacion, desarrollo",,,"foo,bar",,,

This approach works because everything up to the first " is field 1, and everything from there to the second " is field 2 and so on so everything between "s is the even-numbered fields. It can only fail if you have newlines or escaped double quotes inside your fields but that'd affect every other possible solution too so you'd need to add cases like that to your sample input if you want a solution that handles it.