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

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


regex bash shell search replace

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

regex bash shell search replace

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"  fidone < "$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,yyyyyyyyyyyyyxxxxxxxxx,2011-04-10,yyyyyyyyyyyyyxxxxxxxxx,2012-05-10,yyyyyyyyyyyyyxxxxxxxxx,2011-06-10,yyyyyyyyyyyyy

regex bash shell search replace

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/'

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