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digitalmars.D.learn - Differences between two dates (in days...)

reply Mil58 <peter_pinkness yahoo.fr> writes:
Hi All...
I am desperate for the answer to the following problem:
to obtain the difference between the date of today and an older 
date (results in days...)

See my script below, where I would like to do:
"date of today" (var: auj) - "an older date" (var: deb) = xx days

import std.stdio;
import std.datetime;
import core.time : Duration;

void main()
{
auto deb = DateTime(2019, 9, 5);
auto auj = Clock.currTime();
writeln("Aujourd'hui : ", auj.day, " ", auj.month, " ", auj.year);
writeln("Le début : ", deb.day, " ", deb.month, " ", deb.year);
}

Thanks in advance !  :-)
Oct 24 2019
next sibling parent reply Adam D. Ruppe <destructionator gmail.com> writes:
On Thursday, 24 October 2019 at 11:50:04 UTC, Mil58 wrote:

 See my script below, where I would like to do:
 "date of today" (var: auj) - "an older date" (var: deb) = xx 
 days


Try this:

Duration diff = cast(DateTime) auj - deb;
writeln("Diff : ", diff.total!"days");


So basically, just subtract them. You'll need them to both be 
SysTime (which has a timezone) or DateTime (which does not) - if 
you need to convert one, just use cast like I did there so they 
match.

The subtraction will return a Duration object. This has several 
methods to get units, or you can compare it directly

writeln(diff > 5.days); // legal

or like I did there, the `total` method gives the total number of 
units.

see the docs here

http://dpldocs.info/experimental-docs/core.time.Duration.html

and specifically for these methods there are examples which may 
be easier to read than the docs by themselves

http://dpldocs.info/experimental-docs/core.time.Duration.total.html
http://dpldocs.info/experimental-docs/core.time.Duration.split.html
Oct 24 2019
parent Mil58 <peter_pinkness yahoo.fr> writes:
On Thursday, 24 October 2019 at 12:01:21 UTC, Adam D. Ruppe wrote:

 On Thursday, 24 October 2019 at 11:50:04 UTC, Mil58 wrote:

 [...]


 Try this:

 Duration diff = cast(DateTime) auj - deb;
 writeln("Diff : ", diff.total!"days");

 [...]


Awsome ! Thank you for your quick reply...
Both lines work perfectly  :-)
Oct 24 2019
prev sibling parent Daniel Kozak <kozzi11 gmail.com> writes:
On Thu, Oct 24, 2019 at 1:55 PM Mil58 via Digitalmars-d-learn
<digitalmars-d-learn puremagic.com> wrote:

 Hi All...
 I am desperate for the answer to the following problem:
 to obtain the difference between the date of today and an older
 date (results in days...)

 See my script below, where I would like to do:
 "date of today" (var: auj) - "an older date" (var: deb) = xx days

 import std.stdio;
 import std.datetime;
 import core.time : Duration;

 void main()
 {
 auto deb = DateTime(2019, 9, 5);
 auto auj = Clock.currTime();
 writeln("Aujourd'hui : ", auj.day, " ", auj.month, " ", auj.year);
 writeln("Le début : ", deb.day, " ", deb.month, " ", deb.year);
 }

 Thanks in advance !  :-)


import std.stdio;
import std.datetime;
import core.time : Duration;

void main()
{
auto deb = DateTime(2019, 9, 5);
auto auj = Clock.currTime();
writeln("Aujourd'hui : ", auj.day, " ", auj.month, " ", auj.year);
writeln("Le début : ", deb.day, " ", deb.month, " ", deb.year);
writeln("Diff in days : ", (cast(DateTime)auj - deb).total!"days");
}
Oct 24 2019