Bash Script Regular Expressions…How to find and replace all matches?

Question

I am writing a bash script that reads a file line by line.

The file is a .csv file which contains many dates in the format DD/MM/YYYY but I would like to change the

Answer 1

Pure Bash.

infile='data.csv'

while read line ; do
  if [[ $line =~ ^(.*),([0-9]{1,2})/([0-9]{1,2})/([0-9]{4}),(.*)$ ]] ; then
    echo "${BASH_REMATCH[1]},${BASH_REMATCH[4]}-${BASH_REMATCH[3]}-${BASH_REMATCH[2]},${BASH_REMATCH[5]}"
  else
    echo "$line"
  fi
done < "$infile"

The input file

xxxxxxxxx,11/03/2011,yyyyyyyyyyyyy          
xxxxxxxxx,10/04/2011,yyyyyyyyyyyyy          
xxxxxxxxx,10/05/2012,yyyyyyyyyyyyy          
xxxxxxxxx,10/06/2011,yyyyyyyyyyyyy          

gives the following output:

xxxxxxxxx,2011-03-11,yyyyyyyyyyyyy
xxxxxxxxx,2011-04-10,yyyyyyyyyyyyy
xxxxxxxxx,2012-05-10,yyyyyyyyyyyyy
xxxxxxxxx,2011-06-10,yyyyyyyyyyyyy