;-)
Thanx Peter. I'm too not thought about solutions :-).
I was so focused to mined data from a CSV file, and I've not think much
about the appropriate algorithm.
I love attentive and thoughtful people (although often I'm not alone such)
Maybe I corrected in a subsequent e-mail properly.
One
Hi Pablo
Here is a quick solution. However, it not treat the incorrect input
data, ie it requires, however, correct input, otherwise it will collapse ...
Jaroslav Hajtmar
\usemodule[scancsv]
\def\ddmm#1#2#3{% #1 - date, #2 - old separator, #3 - new separator
\startlua
Hi Pablo,
I am sending still slightly modified version (use local variables and
better typographic output).
It use of much variables, but at least it is clear how it works. For
solution occurred me to use parsing function that I have been defined in
the library, so that it can be successfully
On Thu, Dec 18 2014, Jaroslav Hajtmar wrote:
if (year2000) then year=year+2000 end;
Hi,
What about Test;11/11/1999 ... ;-)
--
Peter
___
If your question is of interest to others as well, please add
On 12/18/2014 01:24 PM, Jaroslav Hajtmar wrote:
Hi Pablo,
I am sending still slightly modified version (use local variables and
better typographic output).
It use of much variables, but at least it is clear how it works. For
solution occurred me to use parsing function that I have been
On 12/18/2014 01:37 PM, Peter Münster wrote:
On Thu, Dec 18 2014, Jaroslav Hajtmar wrote:
if (year2000) then year=year+2000 end;
Hi,
What about Test;11/11/1999 ... ;-)
Hi Peter,
all dates start from this year :-).
Pablo
--
http://www.ousia.tk
Hi Pablo.
Peter pointed out an error in my solution. Thanks Peter! His message but
pointed out that in fact is not the task as simple as it seems at first
glance. In fact, a lot depends on what target group for those you solve
a problem resp. whether e.g. date of birth of persons. What does it
