Question

所以我有桌子筛选：

Name                                      Null?      Type
--------------------------------------- --------  -------------

 SCREENINGNO                             NOT NULL   VARCHAR2(8)

 CINEMANO                                           VARCHAR2(8)

 MOVIENO                                            VARCHAR2(8)

 SUPERVISORNO                                       VARCHAR2(8)

 CAMERAOPNO                                         VARCHAR2(8)

 HOURSTART                                          NUMBER(38)

 SCREENINGDATE                                      DATE

问题是＆＃34;显示2015年每个月安排的放映次数。＆＃34;

我试过了：

SELECT to_char(s1.screeningdate, 'mm') AS month
    ,count(s2.screeningno) AS number_of_times_screened
FROM screening s1
    ,screening s2
WHERE s1.movieno(+) = s2.movieno
    AND to_char(s1.screeningdate, 'yy') = '15'
GROUP BY to_char(s1.screeningdate, 'mm')
ORDER BY to_char(s1.screeningdate, 'mm');

它返回了：

MO            NUMBER_OF_TIMES_SCREENED
--            ------------------------ 
01                       22

02                       31

03                       24

04                       32

05                       26

06                       37

07                       15

08                       15

10                       10

11                        5

12                        8

选择了11行。

它只返回11行，我需要它来显示9月。有人可以帮忙吗？

Answer 1

这样做的一种方法是使用带有dual子句的connect by level psedotable来生成1到12之间的所有数字，并将其加入到筛选计数中：

SELECT    d.month, cnt
FROM      (SELECT     LPAD(LEVEL, 2, '0') AS month
           FROM       dual
           CONNECT BY LEVEL <= 12) d
LEFT JOIN (SELECT   TO_CHAR(screeningdate, 'mm') AS month, COUNT(*) AS cnt
           FROM     screeninginfo
           WHERE    TO_CHAR(screeningdate, 'yy') = '15'
           GROUP BY TO_CHAR(screeningdate, 'mm')) s ON d.month = s.month
ORDER BY  1 ASC

当count使用1表返回null时显示所有行

1 个答案: