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 ?
Using EasyPattern for capture ?
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Re: Using EasyPattern for 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 ) ])
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 ) ])