如何从 Oracle 程序中获取单个结果？

发布于 2024-09-13 00:50:30 字数 536 浏览 3 评论 0原文

我有这个程序：

CREATE OR REPLACE PROCEDURE CONV1(
    pDate   IN  VARCHAR2,
    pYear  OUT number,
    pMonth OUT number,
    pDay   OUT number
)
AS
    lDate   DATE;
BEGIN
    lDate := to_date(pDate, 'DD.MM.YYYY HH24:MI:SS');
    pYear := to_number(to_char(lDate, 'YYYY'));
    pMonth := to_number(to_char(lDate, 'MM'));
    pDay := to_number(to_char(lDate, 'DD'));

END CONV1;
/

如果我只想要其中一个出局，我该如何调用这个程序？（就像Select FMAN_STAT_CONV1('16.07.2010', pDay) from Dual;（顺便说一句，这是行不通的））

再见！

原文

I've got this procedure:

CREATE OR REPLACE PROCEDURE CONV1(
    pDate   IN  VARCHAR2,
    pYear  OUT number,
    pMonth OUT number,
    pDay   OUT number
)
AS
    lDate   DATE;
BEGIN
    lDate := to_date(pDate, 'DD.MM.YYYY HH24:MI:SS');
    pYear := to_number(to_char(lDate, 'YYYY'));
    pMonth := to_number(to_char(lDate, 'MM'));
    pDay := to_number(to_char(lDate, 'DD'));

END CONV1;
/

How do I call this procedure if I just want ONE of the outs in there?
(Like Select FMAN_STAT_CONV1('16.07.2010', pDay) from dual; (which ain't work btw))

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み格子的夏天 2024-09-20 00:50:30

创建将使用过程 conv1 的函数，但仅返回一个值。

或者更适合您的特定情况

SELECT to_char(to_date(your_date, 'DD.MM.YYYY HH24:MI:SS'), 'DD') from dual.

或常见情况是：

CREATE OR REPLACE FUNCTION CONV2(
  pDate   IN  VARCHAR2
) 
RETURN NUMBER
IS
  pDay   number;
  pMonth number;
  pYear  number;

BEGIN
   conv1(pDate, pYear, pMonth, pDay);
   return pDay;
END;

Create function which will use procedure conv1, but will return only one value.

Or even better for your particular case

SELECT to_char(to_date(your_date, 'DD.MM.YYYY HH24:MI:SS'), 'DD') from dual.

Or common case is:

CREATE OR REPLACE FUNCTION CONV2(
  pDate   IN  VARCHAR2
) 
RETURN NUMBER
IS
  pDay   number;
  pMonth number;
  pYear  number;

BEGIN
   conv1(pDate, pYear, pMonth, pDay);
   return pDay;
END;

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