Wolfram mathematica 选择SQLDateTimes的范围
我想返回Wolfram mathematica 选择SQLDateTimes的范围,wolfram-mathematica,date-range,sqldatetime,Wolfram Mathematica,Date Range,Sqldatetime,我想返回表中SQLDateTime在上一年内的所有案例(DatePlus[{-1,“year”}])。如何指定这些案例的搜索 您也可以使用日期差异: table = {{ID1, SQLDateTime[{1978, 1, 10, 0, 0, 0.`}]}, {ID2, SQLDateTime[{1999, 1, 10, 0, 0, 0.`}]}, {ID3, SQLDateTime[{2010, 9, 10, 0, 0, 0.`}]}, {ID4, SQLDateTime[{2011,
表
中SQLDateTime在上一年内的所有案例(DatePlus[{-1,“year”}]
)。如何指定这些案例的搜索 您也可以使用日期差异:
table = {{ID1, SQLDateTime[{1978, 1, 10, 0, 0, 0.`}]},
{ID2, SQLDateTime[{1999, 1, 10, 0, 0, 0.`}]},
{ID3, SQLDateTime[{2010, 9, 10, 0, 0, 0.`}]},
{ID4, SQLDateTime[{2011, 1, 10, 0, 0, 0.`}]}}
Cases[table,{a_,SQLDateTime[b_]}/;
DateDifference[b,DateList[],“Year”[[1]]Select[table,(AbsoluteTime[DatePlus[{-1,“Year”}]]]@Leonid;-D为未来干杯dates@Sjoerd是的,我要为此干杯:)
Cases[table, {a_, SQLDateTime[b_]} /;
DateDifference[b, DateList[], "Year"][[1]] <= 1]
Select[table, (AbsoluteTime[ DatePlus[{-1, "Year"}]] <=
AbsoluteTime[ #[[2, 1]]] <= AbsoluteTime[ ] &)]
(* ==> {{ID3, SQLDateTime[{2010, 9, 10, 0, 0, 0.}]},
{ID4,SQLDateTime[{2011, 1, 10, 0, 0, 0.}]}
}
*)
With[
{date = Date[]},
Select[table,
(AbsoluteTime[ DatePlus[date, {-1, "Year"}]] <=
AbsoluteTime[ #[[2, 1]]] <= AbsoluteTime[date ] &)]
]