Question

拥有这3张桌子

第一个，EMPLOYEE

//Employee
E#     ENAME         D#
-----------------------
1      ALI           1
2      SITI          2
3      JOHN          2
4      MARY          3
5      CHIRS         3

第二，PROJECT

//PROJECT
P#     PNAME        D#
-----------------------
1     Computing     1
2     Coding        3
3     Researching   3

第三，WORKSON

//WORKSON
E#     P#     Hours
--------------------
1      1       3
1      2       5
4      3       6

所以我的输出应该是

E#       ENAME      D#       TOTAL HOURS/W
--------------------------------------------
1         ALI       1              8
2        SITI       2              0
3        JOHN       2              0
4         MAY       3              6
5        CHIRS      3              0

显示0，因为员工没有可以处理的项目。

我目前使用

的陈述

SELECT E.E#,E.NAME,E.D#,NVL(sum(Hours),0) AS "TOTAL HOURS/W"
    FROM Employee E,WORKSON W WHERE E.E#=W.E# OR 
GROUP BY E.E#,E.NAME,E.D#;

不知道如何选择然后将显示所有员工，而不仅仅是特定的E＃。输出显示为员工没有任何项目工作，应显示0值

Answer 1

这可以通过

加入员工和WORKSON来完成

select
e.`E#`,
e.ENAME,
e.`D#`,
coalesce(tot,0) as `TOTAL HOURS/W`
from Employee e
left join
(
  select `E#`,
  sum(Hours) as tot
  from WORKSON
  group by `E#`
)w
on w.`E#` = e.`E#`
group by e.`E#`

<强> DEMO

Answer 2

您应该使用left outer join。以下内容适用于MySQL和Oracle：

SELECT E.E#, E.NAME, E.D#,
       COALESCE(SUM(Hours), 0) AS "TOTAL HOURS/W"
FROM Employee E LEFT OUTER JOIN
     WORKSON W 
     ON E.E# = W.E# 
GROUP BY E.E#, E.NAME, E.D#;

带有SELECT子句的MYSQL有三个不同的表

2 个答案: