Re: A Question Derived from Converting decimal value

[ThePorgie]

Thank you sir! Don't know how I missed that in the manual and I promise I 
did look there first.

On Wednesday, March 16, 2022 at 4:13:30 PM UTC-4 Patrick Woolsey wrote:

> On 3/16/22 at 2:06 PM, thep...@gmail.com (ThePorgie) wrote:
>
> >I was tinkering with with Patrick's question
>
> Not actually my question ;-) though close enough for the matters 
> at hand.
>
>
> >So I was looking at this and knew it would have to be several 
> >passes, but my pattern I was working with for the first pass 
> >was "(?P\d+\.)1"
> >
> >My question is since I can't use a replace pattern \1 follow by 33
>
> You can in fact do this by prefixing a zero "0" to the singular 
> backreference number, so:
>
> Replace: \0133
>
> will give you the first backreference \01 followed by the string "33".
>
>
> >I was using a named pattern, but (?P=foo) doesn't call the named pattern
> >in the replace area of a grep search? Just curious as I couldn't find
> >anything regarding this in the manual.
>
> Please see the section titled "Subpatterns Make Replacement 
> Powerful" in Chapter 8 (page 202) of the current manual:
>
> Pattern Inserts
> =
> [...]
>
> \P the text matched by the subpattern NAME
>
>
> Regards,
>
> Patrick Woolsey
> ==
> Bare Bones Software, Inc. 
>
>

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Re: A Question Derived from Converting decimal value

[Patrick Woolsey]

On 3/16/22 at 2:06 PM, thepo...@gmail.com (ThePorgie) wrote:


I was tinkering with with Patrick's question


Not actually my question ;-) though close enough for the matters 
at hand.



So I was looking at this and knew it would have to be several 
passes, but my pattern I was working with for the first pass 
was "(?P\d+\.)1"


My question is since I can't use a replace pattern \1 follow by 33


You can in fact do this by prefixing a zero "0" to the singular 
backreference number, so:


Replace:\0133

will give you the first backreference \01 followed by the string "33".



I was using a named pattern, but (?P=foo) doesn't call the named pattern
in the replace area of a grep search? Just curious as I couldn't find
anything regarding this in the manual.


Please see the section titled "Subpatterns Make Replacement 
Powerful" in Chapter 8 (page 202) of the current manual:


Pattern   Inserts
=
[...]

\P  the text matched by the subpattern NAME


Regards,

 Patrick Woolsey
==
Bare Bones Software, Inc. 

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A Question Derived from Converting decimal value

[ThePorgie]

I was tinkering with with Patrick's question regarding replacing the digits 
where the "n.1" is replaced with "n.33" and "n.2" is replaced with "n.67"
His sample data:
*Sample  Desired output*
0.20.67
1.11.33
1.11.33
1.21.67
1.11.33
66
3.2 3.67
88
6.2 6.67

So I was looking at this and knew it would have to be several passes, but 
my pattern I was working with for the first pass was "(?P\d+\.)1"
My question is since I can't use a replace pattern \1 follow by 33 I was 
using a named pattern, but (?P=foo) doesn't call the named pattern in the 
replace area of a grep search? Just curious as I couldn't find anything 
regarding this in the manual.

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