3

为什么我的 PL/SQL 持续时间函数不能正常工作？在下面的查询中，我以与下面的函数相同的方式手动计算“hh:mm”。但是，我得到不同的结果。

来电查询：

WITH durdays AS (SELECT SYSDATE - (SYSDATE - (95 / 1440)) AS durdays FROM DUAL)
SELECT TRUNC (24 * durdays) AS durhrs,
       MOD (TRUNC (durdays * 1440), 60) AS durmin,
       steven.duration (SYSDATE - (95 / 1400), SYSDATE) duration
  FROM durdays

输出：

durhrs:   1
durmin:   35
duration: '1:38'

功能代码：

CREATE OR REPLACE FUNCTION steven.duration (d1 IN DATE, d2 IN DATE)
   RETURN VARCHAR2 IS
   tmpvar   VARCHAR2 (30);
   durdays   NUMBER (20,10); -- Days between two DATEs
   durhrs   BINARY_INTEGER; -- Completed hours
   durmin   BINARY_INTEGER; -- Completed minutes
BEGIN
   durdays   := d2-d1; 
   durhrs   := TRUNC(24 * durdays);   
   durmin   := MOD (TRUNC(durdays * 1440), 60);
   tmpvar   := durhrs || ':' || durmin;
   RETURN tmpvar;
END duration;
/
4

2 回答 2

5

我想你可能有一个小错字：

WITH durdays AS (SELECT SYSDATE - (SYSDATE - (95 / 1440)) AS durdays FROM DUAL)
                                                   ^^^^
SELECT TRUNC (24 * durdays) AS durhrs,
       MOD (TRUNC (durdays * 1440), 60) AS durmin,
       lims_sys.duration (SYSDATE - (95 / 1400), SYSDATE) duration
                                          ^^^^

一旦解决了这个问题，它就可以正常工作：

SQL> WITH durdays AS (SELECT SYSDATE - (SYSDATE - (95 / 1440)) AS durdays FROM DUAL)
  2  SELECT TRUNC (24 * durdays) AS durhrs,
  3         MOD (TRUNC (durdays * 1440), 60) AS durmin,
  4         duration (SYSDATE - (95 / 1440), SYSDATE) duration
  5    FROM durdays
  6  ;

    DURHRS     DURMIN DURATION
---------- ---------- ----------------
         1         34 1:34
于 2009-09-15T15:40:20.717 回答
-1 
sql>set time on

然后你会得到如下提示：

20:24:35 sql> 从 v$session 中选择计数（*）；

数数（*）
------------
78

20:24:50 sql>

现在你可以知道查询持续时间是 15 秒

于 2011-12-06T13:19:39.083 回答