DateConvert
Synopsis
DateConvert(date,vbToInternal) DateConvert(date,vbToExternal)
Arguments
| date | The date to be converted. An external date is represented as a string, such as “10–22–1980”. An internal date is represented as a five-digit integer, which is the first part of the Caché $HOROLOG ($H) date/time format. |
| vbToInternal | This keyword specifies converting an external date to internal ($H) format. |
| vbToExternal | This keyword specifies converting an internal date ($H format) to external date format. |
Description
If you specify a date/time value, DateConvert ignores the time portion.
The DateConvert function returns an external date in the following format:
mm/dd/yyyy
Leading zeros are displayed. The year is displayed as four digits.
The DateConvert function returns an internal date/time in the following format:
ddddd
Where “d” is the date count (number of days since 12/31/1840). For further details, see $HOROLOG in the Caché ObjectScript Reference.
An omitted year value defaults to 2000; the two-digit year defaults are 2000 through 2029 (for 00 through 29) and 1930 through 1999 (for 30 through 99).
Examples
The following example takes an external date/time value, converts the date part to an internal format ($HOROLOG) value, then converts this internal value back to an external format date.
Dim GetDate,InDate, ExDate
GetDate = "1-12-1953 11:45:23"
Println GetDate
InDate = DateConvert(GetDate,vbToInternal)
Println InDate
ExDate = DateConvert(InDate,vbToExternal)
Println ExDateThe values printed are as follows:
1-12-1953 11:45:23 40919 01/12/1953
See Also
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Basic: DateTimeConvert function
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Basic: TimeConvert function
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ObjectScript: $HOROLOG special variable
