我想显示下面给定表格的数据透视表(交叉表)。
表:Employee
CREATE TABLE Employee
(
Employee_Number varchar(10),
Employee_Role varchar(50),
Group_Name varchar(10)
);
插入:
INSERT INTO Employee VALUES('EMP101','C# Developer','Group_1'),
('EMP102','ASP Developer','Group_1'),
('EMP103','SQL Developer','Group_2'),
('EMP104','PLSQL Developer','Group_2'),
('EMP101','Java Developer',''),
('EMP102','Web Developer','');
现在我想显示上述数据的数据透视表,如下所示:
预期结果:
Employee_Number TotalRoles TotalGroups Available Others Group_1 Group_2
---------------------------------------------------------------------------------------------------
EMP101 2 2 1 1 1 0
EMP102 2 2 1 1 1 0
EMP103 1 2 1 0 0 1
EMP104 1 2 1 0 0 1
解释:我想展示每位员工所拥有的Employee_Number
,TotalRoles
,
TotalGroups
显示给所有员工的Available
,显示员工可用
在多少组中,Others
必须显示该员工在其他组中也可用
group_name尚未分配,最后Group_Names
必须以数据透视表格式显示。
答案 0 :(得分:1)
您可以使用crosstab功能。
首先,如果您还没有添加tablefunc扩展名:
CREATE EXTENSION tablefunc;
交叉表函数要求您传递一个查询,返回您需要转动的数据,然后输出输出中的列列表。 (以其他方式"告诉我你想要的输入和输出格式")。排序顺序很重要!
在您的情况下,输入查询非常复杂 - 我认为您需要加载单独的查询,然后UNION ALL它们以获取所需的数据。我不完全确定你如何计算价值" TotalGroups"和"可用",但您可以在相关位置修改以下内容以获得所需内容。
SELECT * FROM crosstab(
'SELECT employee_number, attribute, value::integer AS value FROM (with allemployees AS (SELECT distinct employee_number FROM employee) -- use a CTE to get distinct employees
SELECT employee_number,''attr_0'' AS attribute,COUNT(distinct employee_role) AS value FROM employee GROUP BY employee_number -- Roles by employee
UNION ALL
SELECT employee_number,''attr_1'' AS attribute,value from allemployees, (select count (distinct group_name) as value from employee where group_name <> '''') a
UNION ALL
SELECT employee_number,''attr_2'' AS attribute, COUNT(distinct group_name) AS value FROM employee where group_name <> '''' GROUP BY employee_number -- Available, do not know how this is calculate
UNION ALL
SELECT a.employee_number, ''attr_3'' AS attribute,coalesce(value,0) AS value FROM allemployees a LEFT JOIN -- other groups. Use a LEFT JOIN to avoid nulls in the output
(SELECT employee_number,COUNT(*) AS value FROM employee WHERE group_name ='''' GROUP BY employee_number) b on a.employee_number = b.employee_number
UNION ALL
SELECT a.employee_number, ''attr_4'' AS attribute,coalesce(value,0) AS value FROM allemployees a LEFT JOIN -- group 1
(SELECT employee_number,COUNT(*) AS value FROM employee WHERE group_name =''Group_1'' GROUP BY employee_number) b on a.employee_number = b.employee_number
UNION ALL
SELECT a.employee_number, ''attr_5'' AS attribute,coalesce(value,0) AS value FROM allemployees a LEFT JOIN -- group 2
(SELECT employee_number,COUNT(*) AS value FROM employee WHERE group_name =''Group_2'' GROUP BY employee_number) b on a.employee_number = b.employee_number) a order by 1,2')
AS ct(employee_number varchar,"TotalRoles" integer,"TotalGroups" integer,"Available" integer, "Others" integer,"Group_1" integer, "Group_2" integer)
答案 1 :(得分:1)
SELECT * FROM crosstab(
$$SELECT grp.*, e.group_name
, CASE WHEN e.employee_number IS NULL THEN 0 ELSE 1 END AS val
FROM (
SELECT employee_number
, count(employee_role)::int AS total_roles
, (SELECT count(DISTINCT group_name)::int
FROM employee
WHERE group_name <> '') AS total_groups
, count(group_name <> '' OR NULL)::int AS available
, count(group_name = '' OR NULL)::int AS others
FROM employee
GROUP BY 1
) grp
LEFT JOIN employee e ON e.employee_number = grp.employee_number
AND e.group_name <> ''
ORDER BY grp.employee_number, e.group_name$$
,$$VALUES ('Group_1'::text), ('Group_2')$$
) AS ct (employee_number text
, total_roles int
, total_groups int
, available int
, others int
, "Group_1" int
, "Group_2" int);
SQL Fiddle演示了基本查询,但没有安装在sqlfiddle.com上的交叉表步骤
交叉表的基础知识:
这个交叉表中的特别之处:所有&#34;额外&#34;列。这些列位于&#34;行名称&#34;之后的中间。但之前&#34;类别&#34;和&#34;价值&#34;:
再一次,如果你有一组动态的组,你需要动态构建这个语句并在第二次调用中执行它: