進行中の 4gl の月の最終日を取得するにはどうすればよいですか?
12857 次
10 に答える
13
/* the last day of this month is one day less than the first day of next month
*
* so add one month to the first day of this month and then subtract one day.
*
*/
function lastDay returns date ( input d as date ):
return add-interval( date( month( d ), 1, year( d )), 1, "month" ) - 1.
end.
于 2012-06-19T07:39:36.833 に答える
1
DEF VAR dt-ref AS DATE NO-UNDO.
DEF VAR dt-end-of-month AS DATE NO-UNDO.
ASSIGN dt-ref = DATE(2,12,2012)
.
ASSIGN dt-ref = dt-ref + 33
dt-end-of-month = DATE(MONTH(dt-ref),1,YEAR(dt-ref)) - 1.
于 2012-08-09T14:15:16.423 に答える
1
プラットフォームに「間隔」がない場合 - これも同様に機能します。
DEFINE VARIABLE start-date AS DATE NO-UNDO.
DEFINE VARIABLE end-of-month AS DATE NO-UNDO.
ASSIGN
start-date = DATE(2, 15, 2012)
.
ASSIGN
end-of-month = DATE(MONTH(start-date), 20, YEAR(start-date)) + 15
end-of-month = end-of-month - DAY(end-of-month)
.
于 2012-06-19T13:23:21.490 に答える
0
最も簡単な方法は、ADD-INTERVAL 関数を使用することです。
ldEndDate = ADD-INTERVAL(ldBeginDate, 1, "months") - 1.
于 2014-07-01T07:44:31.347 に答える
0
これはすべきです。
DEFINE VARIABLE InputDate AS DATE NO-UNDO.
DEFINE VARIABLE OutputDate AS DATE NO-UNDO.
IF MONTH(InputDate) = 12 THEN
ASSIGN OutputDate = DATE(1,1,(YEAR(InputDate) + 1 ) ) - 1 .
ELSE
ASSIGN OutputDate = DATE((MONTH(InputDate) + 1 ), 1,YEAR(InputDate) ) - 1.
于 2018-04-06T11:55:04.423 に答える
0
PROCEDURE getEndOfMonth:
DEFINE INPUT PARAMETER lv-date AS DATE.
DEFINE OUTPUT PARAMETER lv-monthEnd AS DATE.
lv-monthEnd = DATE(MONTH(lv-date), 31, YEAR(lv-date)) NO-ERROR.
IF lv-monthEnd = ? THEN DO:
lv-monthEnd = DATE(MONTH(lv-date), 30, YEAR(lv-date)) NO-ERROR.
IF lv-monthEnd = ? THEN DO:
lv-monthEnd = DATE(MONTH(lv-date), 29, YEAR(lv-date)) NO-ERROR.
IF lv-monthEnd = ? THEN
lv-monthEnd = DATE(MONTH(lv-date), 28, YEAR(lv-date)) NO-ERROR.
END.
END.
END.
于 2012-06-19T04:29:50.107 に答える