Nice solution.. very elegant and simple. Thanks, I’ll use this from now on!

Cheers,

Dave

> On 3-October-2019, at 8:16 PM, Bruce Van Allen <[email protected]> wrote:
> 
> On 10/2/19 at 5:01 AM, [email protected] (David Rostenne) wrote:
> 
>> I was hoping for an alternative to the & in the replacement strung that 
>> means ‘entire source line’ but this works perfectly.
> 
> Surround the whole pattern with parentheses. That will then be the capture 
> contained in \1. You will then have to increment each other capture by 1.
> 
> Original:
> http://ftp.newedinburgh.ca/wp-content/uploads/2019/09/1976_04-April-New-Edinburgh-News_web.pdf
> 
> Pattern:
> (.+?/2019/[019]+/([0-9]+)[_.]([0-9]+)-([a-zA-Z]+).*\n)
> 
> Replacement:
> \4 \2, NEN PDF, \2/\3/01, \1
> 
> Result:
> April 1976, NEN PDF, 1976/04/01, 
> http://ftp.newedinburgh.ca/wp-content/uploads/2019/09/1976_04-April-New-Edinburgh-News_web.pdf
> 
> Note that the above pattern includes the end of line ("\n"); if you don't 
> want that, move the closing parenthesis to before the "\n".
> 
> HTH
> -- 
> 
>  - Bruce
> 
> _bruce__van_allen__santa_cruz__ca_
> 
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