Oracle 持续时间函数答案

【问题标题】：Oracle Duration FunctionOracle 持续时间函数
【发布时间】：2010-11-28 12:19:50
【问题描述】：

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，我得到了不同的结果。

调用查询：

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;
/

【问题讨论】：

标签： sql oracle plsql duration

【解决方案1】：

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
                                          ^^^^

一旦修复，它就可以正常工作了：

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

【讨论】：

只需要第二双眼睛来看看它。谢谢！

【解决方案2】：

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时长是15秒的

【讨论】：

这个问题的答案是什么？