Question

我有下表。

CREATE TABLE Employee_id_credits ( Employee_id, credits ) AS
  SELECT 10, 1 FROM DUAL UNION ALL
  SELECT 12, 1 FROM DUAL UNION ALL
  SELECT 10, 1 FROM DUAL UNION ALL
  SELECT 12, 1 FROM DUAL UNION ALL
  SELECT 12, 1 FROM DUAL UNION ALL
  SELECT 14, 1 FROM DUAL;

以下查询对员工的信用总数进行分组和计算。

select Employee_id, count(*) as "Total_credits"
from Employee_id_credits
group by Employee_id;

给出以下输出。

Employee_id Total_credits
----------- -------------
         10             2
         12             3
         14             1

我有一个带有层次结构的员工经理表。

CREATE TABLE Employee_Manager ( Employee_id, Manager_id ) AS
  SELECT  10, 101  FROM DUAL UNION ALL
  SELECT  12, 120  FROM DUAL UNION ALL
  SELECT  13, 120  FROM DUAL UNION ALL
  SELECT  14, 150  FROM DUAL UNION ALL
  SELECT 101, NULL FROM DUAL UNION ALL
  SELECT 120, 130  FROM DUAL UNION ALL
  SELECT 130, NULL FROM DUAL;

我有一个查询来查找该员工的顶级经理。

SELECT
  Employee_id
FROM 
  Employee_Manager
WHERE
  Manager_id is null
CONNECT BY PRIOR
  Manager_id = Employee_id
START WITH
  Employee_id = '12';

我想结合上面两个查询，以便输出如下所示。如何组合两个查询？

Manager Total_credits
------- -------------
    101             2
    130             3
    150             1

Answer 1

SQL Fiddle

Oracle 11g R2架构设置：

SELECT CONNECT_BY_ROOT( Employee_id ) AS Employee_id,
       COALESCE( manager_id, employee_id ) AS manager_id
FROM   Employee_manager
WHERE  CONNECT_BY_ISLEAF = 1
CONNECT BY PRIOR Manager_id = Employee_id

查询1 - 查找每位员工的经理：

| EMPLOYEE_ID | MANAGER_ID |
|-------------|------------|
|          10 |        101 |
|          12 |        130 |
|          13 |        130 |
|          14 |        150 |
|         101 |        101 |
|         120 |        130 |
|         130 |        130 |

<强> Results ：

SELECT m.manager_id,
       SUM( c.credits ) As total_credits
FROM   Employee_id_credits c
       INNER JOIN
       (
         SELECT CONNECT_BY_ROOT( Employee_id ) AS Employee_id,
                COALESCE( manager_id, employee_id ) AS manager_id
         FROM   Employee_manager
         WHERE  CONNECT_BY_ISLEAF = 1
         CONNECT BY PRIOR Manager_id = Employee_id
       ) m
       ON ( c.employee_id = m.employee_id )
GROUP BY m.manager_id

查询2 - 将其加入信用表并汇总：

| MANAGER_ID | TOTAL_CREDITS |
|------------|---------------|
|        101 |             2 |
|        130 |             3 |
|        150 |             1 |

<强> Results ：

undefined is not an object (evaluating 'navigator.userAgent.indexOf')

<unknown>
     :12:71
loadModuleImplementation
    require.js:213:12
<unknown>
    getScrollPosition.js:12:31

将一个sql查询的结果合并到另一个中

1 个答案: