DataMystic

Hello ,

I have asked you in email , but I am listing it here as well.

I want to change Jan. 18 to (January 18) using easy pattern. You have given me the way to do it by perl that is Jan. (\d+?) , and replace with (january $1)


I wanted to know how would i use easy pattern for the same, i know i can capture it by Jan. [1-2 digit] , but how do i replace it back ?
You have used $1 to store it in perl , how does easy pattern work , and could i get more example of this capture ?

Here's the email reply:

Try EasyPatterns instead and read the help file. The equivalent EasyPattern looks like this:

Jan. ([ capture( 1+ digits ) ])



Replace with is still:

January $1



2. Using EasyPatterns,

(Jan. [ capture( 1+ digits ) ] [ capture( 1+ digits ) ])

Or

(Jan. [ capture( 1+ letters ) ])



3. Use:

[3 lines,

Jan. ([ capture( 1+ digits ) ])