【发布时间】:2021-02-12 20:55:11
【问题描述】:
我正在尝试创建一个查询,我想从一个表中选择员工,从另一个表中计算他们的证书数量,总和,然后显示那些拥有超过 3 个证书的人。我试过了:
SELECT CONCAT (
employee_bt.EMP_FNAME
," "
,employee_bt.EMP_LNAME
) AS "Full Name"
,count(earnedrating_bt.RTG_CODE) AS "Count of Rating"
FROM employee_bt
INNER JOIN earnedrating_bt ON earnedrating_bt.EMP_NUM = employee_bt.EMP_NUM
INNER JOIN rating_bt ON rating_bt.RTG_CODE = earnedrating_bt.RTG_CODE
GROUP BY employee_bt.EMP_NUM
,earnedrating_bt.RTG_CODE
ORDER BY employee_bt.EMP_LNAME ASC
但它会返回类似这样的结果,理想情况下,我只想将这些员工加起来,而不是 3 行,每行 1,我想要 1 行,它只是说 3。
Jeanine Duzak 1
Jeanine Duzak 1
Jeanine Duzak 1
John Lange 1
John Lange 1
John Lange 1
Rhonda Lewis 1
Rhonda Lewis 1
Rhonda Lewis 1
Rhonda Lewis 1
应该变成:
Jeanine Duzak 3
John Lange 3
Rhonda Lewis 4
【问题讨论】:
标签: mysql sql count inner-join aggregate-functions