我有两张桌子：[BRANCH_T]和[EMPLOYEE_T]。
BRANCH_T包含branch_id和manager_emp_id（外键）。
在EMPLOYEE_T中包含emp_id，emp_firstname，emp_lastname，emp_position，emp_gender和branch_id（外键）。
我想展示branch_id，每个分支的经理名称，主管总人数，男性员工总数和女性员工总数。
以下是我的代码：
SELECT b.bch_id AS 'BRANCH NUMBER',
(CASE WHEN emp_position = 'M'
THEN emp_fname + ' ' + emp_lname ELSE '-' END) AS 'MANAGER NAME',
SUM (CASE WHEN e.emp_position = 'S' THEN 1 ELSE 0 END) AS 'TOTAL NUMBER OF SUPERVISOR',
SUM (CASE WHEN e.emp_gender = 'M' THEN 1 ELSE 0 END) AS 'TOTAL MALE EMPPLOYEES',
SUM (CASE WHEN e.emp_gender = 'F' THEN 1 ELSE 0 END) AS 'TOTAL FEMALE EMPLOYEES'
FROM EMPLOYEE_T e, BRANCH_T b
WHERE b.bch_id = e.bch_id
GROUP BY b.bch_id, e.emp_fname, e.emp_lname, e.emp_gender, e.emp_position
这是生成的结果。
+---------------+--------------+-----------------------------+----------------------+------------------------+
| branch number | manager name | total number of supervisors | total male employees | total female employees |
+---------------+--------------+-----------------------------+----------------------+------------------------+
| B001 | EMILIA MOANA | 0 | 0 | 1 |
| B001 | - | 1 | 1 | 0 |
| B001 | - | 0 | 0 | 1 |
| B002 | MARCUS HOW | 1 | 0 | 1 |
+---------------+--------------+-----------------------------+----------------------+------------------------+
我尝试了另一个代码，但是它出错并且无法执行。
SELECT b.bch_id AS 'BRANCH NUMBER',
b.mng_emp_id AS 'MANAGER ID',
(SELECT e.emp_fname + ' ' + e.emp_lname
FROM EMPLOYEE_T e, BRANCH_T b
WHERE b.mng_emp_id = e.emp_id and
b.bch_id = e.bch_id
group by e.emp_fname, e.emp_lname, e.bch_id) AS 'MANAGER NAME',
SUM (CASE WHEN e.emp_position = 'S' THEN 1 ELSE 0 END) AS 'NUMBER OF SUPERVISOR',
SUM (CASE WHEN e.emp_gender = 'M' THEN 1 ELSE 0 END) AS 'NUMBER OF MALE EMPLOYEES',
SUM (CASE WHEN e.emp_gender = 'F' THEN 1 ELSE 0 END) AS 'NUMBER OF FEMALE EMPLOYEES'
FROM EMPLOYEE_T e, BRANCH_T b
WHERE e.bch_id = b.bch_id
GROUP BY b.bch_id, b.mng_emp_id, e.emp_position, e.emp_gender
错误代码：
Subquery returned more than 1 value. This is not permitted when the subquery follows =, !=, <, <= , >, >= or when the subquery is used as an expression.
我想要展示的是这样的：
+---------------+--------------+-----------------------------+----------------------+------------------------+
| branch number | manager name | total number of supervisors | total male employees | total female employees |
+---------------+--------------+-----------------------------+----------------------+------------------------+
| B001 | EMILIA MOANA | 1 | 1 | 2 |
| B002 | MARCUS HOW | 1 | 2 | 1 |
| B003 | ELINA YEAN | 1 | 1 | 1 |
+---------------+--------------+-----------------------------+----------------------+------------------------+
请帮忙。谢谢。
---------------- UPDATE ----------------
我可以成功检索每个分支机构的主管，男性和女性员工的数量，但我只能从BRANCH_T获取mng_emp_id。
SELECT b.bch_id AS 'BRANCH NUMBER',
b.mng_emp_id AS 'MANAGER ID',
SUM (CASE WHEN e.emp_position = 'S' THEN 1 ELSE 0 END) AS 'NUMBER OF SUPERVISOR',
SUM (CASE WHEN e.emp_gender = 'M' THEN 1 ELSE 0 END) AS 'NUMBER OF MALE EMPLOYEES',
SUM (CASE WHEN e.emp_gender = 'F' THEN 1 ELSE 0 END) AS 'NUMBER OF FEMALE EMPLOYEES'
FROM EMPLOYEE_T e, BRANCH_T b
WHERE e.bch_id = b.bch_id
GROUP BY b.bch_id, b.mng_emp_id
mng_emp_id是emp_id的一部分。
BRANCH_T：
+--------+------------+
| BCH_ID | MNG_EMP_ID |
+--------+------------+
| B001 | EM001 |
| B002 | EM002 |
| B003 | EM003 |
+--------+------------+
EMPLOYEE_T：
+--------+-----------+-----------+--------+
| EMP_ID | EMP_FNAME | EMP_LNAME | BCH_ID |
+--------+-----------+-----------+--------+
| EM001 | A | Z | B001 |
| EM002 | B | Y | B002 |
| EM003 | C | X | B003 |
| ES001 | D | W | B001 |
| ET001 | E | V | B002 |
+--------+-----------+-----------+--------+