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

I don't wanna echo...I want to store the result in a new variable

to do that (and i know this isn't exacting to your setup, but still applies in how to use a regex):

path="/entertainment/Pictures"
files=(
  "$path"/*.jpg"
  "$path"/*.bmp"
)

for i in "${files[@]}"
do
  # replace jpg and bmp with png in prep for conversion
  new=$(echo "$i" | perl -pe "s/\.jpg|\.bmp/.png")

  # next is to perform the conversion
  ...
done

Answer 2

Try this using sed:

line='Today is 10/12/2010 and yesterday was 9/11/2010'
echo "$line" | sed -r 's#([0-9]{1,2})/([0-9]{1,2})/([0-9]{4})#\3-\2-\1#g'

OUTPUT:
Today is 2010-12-10 and yesterday was 2010-11-9

PS: On mac use sed -E instead of sed -r

Answer 3

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

Answer 4

You can do it using sed

echo "11/12/2011" | sed -E 's/([0-9][0-9]?)\/([0-9][0-9]?)\/([0-9][0-9][0-9][0-9])/\3-\2-\1/'