日期时间格式我有一列包含几种不同的日期格式，还有一些空白日期。如何修改该列以显示不同日期类型的一种日期列类型？以下是一些值的屏幕截图。
自动化和脚本数据混合和清理回复：日期时间格式这是处理它的一种方法
Names Default To Here( 1 );
dt = New Table( "Example",
Add Rows( 2 ),
New Column( "datetime",
Character,
"Nominal",
Set Values( {"2023-05-10 17:47:36.005390800+00:00", "6/6/2024 9:06"} ),
Set Display Width( 230 )
)
);
New Column( "new datetime", Numeric, "Continuous", Format( "m/d/y h:m:s", 23, 0 ) );
For Each Row(
If( contains(:datetime, "+"),
theDateTime = word(1,:datetime,"+");
substitute into(theDateTime," ","T");
:new datetime = informat(theDateTime,"yyyy-mm-ddThh:mm:ss");
,
contains(:datetime,"/"),
:new datetime = informat(:datetime,"m/d/y h:m")
)
);回复：日期时间格式我会选择和 Jim 类似的方法。使用 If 语句检查当前正在处理的格式，然后使用InFormat()将其转换为数值
/<MM>/<YYYY> <HH24>:<MM>";
);
In Format(
Word(1, :datetime, "."),
"Format Pattern",
pattern
);
)
);
</MM></HH24></YYYY></MM>Names Default To Here(1);
dt = New Table("Example",
Add Rows(2),
New Column("datetime",
Character,
"Nominal",
Set Values({"2023-05-10 17:47:36.005390800+00:00", "6/6/2024 9:06"})
)
);
// If you wish to keep milliseconds some extra stuff is needed
dt << new column("date", numeric, continuous, Format("Locale Date Time h:m:s", 23, 0),
Formula(
If(Contains(:datetime, "+"),
pattern = "<YYYY>-<MM>-<DL><DD> <HH24>:<MM>:<SS>";
,
pattern = "</SS></MM></HH24></DD></DL></MM></YYYY>
