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Considering a DateObject d and TimeObject t:

In[1]:= t = TimeObject[{14, 1, 45.`}]
In[2]:= d = DateObject[{2012, 06, 11}]

How can I achieve this result:

In[3]:= dt = DateObject[{2012, 06, 11, 14, 1, 45.`}]

Thanks!

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Another, shorter way is

Append[d, t]

Mathematica graphics

From the docs,

DateObject[date,time] represents the specified date list and TimeObject time.

If you need the list you mention in the question, just convert the DateObject using DateList:

DateList@Append[d, t]
(* {2012, 6, 11, 14, 1, 45.} *)
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A little late to the party here but this also works:

In[28]:= DateObject[d, t]

as Szabolcs noted, the documentation references this, but doesn't give an explicit example(which I will make sure gets added).

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Strip off their heads. replace them with List and then Flatten them:

dt=DateObject@@(List @@@ {d, t} // Flatten)

DateObject[{2012, 6, 11, 14, 1, 45.}]

Or ....

dt=DateObject @@ Join @@ {d[[1]], t[[1]]}
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  • $\begingroup$ Why is the result no quantity/object, but only text? $\endgroup$ – SuTron Jun 17 '15 at 11:06
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    $\begingroup$ My bad, it's early for me ;) $\endgroup$ – image_doctor Jun 17 '15 at 11:09
  • $\begingroup$ Why not Join[] the lists instead? :) $\endgroup$ – J. M.'s technical difficulties♦ Jun 17 '15 at 11:21
  • $\begingroup$ See recent update :) $\endgroup$ – image_doctor Jun 17 '15 at 11:22
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    $\begingroup$ Well, I had DateObject @@ Join[d[[1]], t[[1]]], myself… :) $\endgroup$ – J. M.'s technical difficulties♦ Jun 17 '15 at 11:24
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Well...

d~d[[0]]~t

...seems to work.

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