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vi command for adding character
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geoffhazel
Mar 19, 2012, 12:49:45 PM3/19/12
to
I have a file like this
02/1/2012
02/2/2012
02/3/2012
03/1/2012
03/2/2012
03/3/2012
I want it to be 02/01/2012 and so forth, MM/DD/YYYY format with extra
0's as placeholders.
I can change all the /1/ to /01/ with a command :%s?/1?/01? but what
I want is to be able to do is to insert a 0 in front of all the cases
where /./ happens. I can find the patterns with /./ but how do I
preserve the characters that are there? I seem to recall there's a way
to specify those with some regex but can't recall the details.
02/1/2012
02/2/2012
02/3/2012
03/1/2012
03/2/2012
03/3/2012
I want it to be 02/01/2012 and so forth, MM/DD/YYYY format with extra
0's as placeholders.
I can change all the /1/ to /01/ with a command :%s?/1?/01? but what
I want is to be able to do is to insert a 0 in front of all the cases
where /./ happens. I can find the patterns with /./ but how do I
preserve the characters that are there? I seem to recall there's a way
to specify those with some regex but can't recall the details.
Ivan Shmakov
Mar 19, 2012, 3:04:17 PM3/19/12
to
>>>>> geoffhazel <geoff...@gmail.com> writes:
[Cross-posting to news:comp.editors and setting Followup-To:
there, for the question is hardly Unix-specific.]
> I have a file like this
> 02/1/2012
> 02/2/2012
> 02/3/2012
> 03/1/2012
> 03/2/2012
> 03/3/2012
> I want it to be 02/01/2012 and so forth, MM/DD/YYYY format with extra
> 0's as placeholders.
[...]
> I can find the patterns with /./ but how do I preserve the characters
> that are there? I seem to recall there's a way to specify those with
> some regex but can't recall the details.
One can reference a \(\)-group from within the s-command's
replacement text with \1 ... \9, like:
s;/\([0-9]/\);/0\1;
--
FSF associate member #7257
[Cross-posting to news:comp.editors and setting Followup-To:
there, for the question is hardly Unix-specific.]
> I have a file like this
> 02/1/2012
> 02/2/2012
> 02/3/2012
> 03/1/2012
> 03/2/2012
> 03/3/2012
> I want it to be 02/01/2012 and so forth, MM/DD/YYYY format with extra
> 0's as placeholders.
> I can find the patterns with /./ but how do I preserve the characters
> that are there? I seem to recall there's a way to specify those with
> some regex but can't recall the details.
replacement text with \1 ... \9, like:
s;/\([0-9]/\);/0\1;
--
FSF associate member #7257
Harry
May 7, 2012, 10:57:34 PM5/7/12
to
geoffhazel wrote...
>
>I have a file like this
>02/1/2012
>02/2/2012
>02/3/2012
>03/1/2012
>03/2/2012
>03/3/2012
>
>I want it to be 02/01/2012 and so forth, MM/DD/YYYY format with extra
>0's as placeholders.
:1,$s#/#/0#
>I have a file like this
>02/1/2012
>02/2/2012
>02/3/2012
>03/1/2012
>03/2/2012
>03/3/2012
>
>I want it to be 02/01/2012 and so forth, MM/DD/YYYY format with extra
>0's as placeholders.
Rui Maciel
May 8, 2012, 5:07:34 AM5/8/12
to
december, that search pattern will cause problems such as:
03/011/2012
The following search pattern won't:
:1,$s#/\([1-9]\/\)#/0\1#
Rui Maciel
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