【发布时间】:2021-03-17 20:10:49
【问题描述】:
我有下表,其中包含 empid、deptid 和名称。我需要得到结果 每个部门的empid,deptid,name和count。
CREATE TABLE emp(empid INTEGER PRIMARY KEY, deptid INTEGER, NAME TEXT);
/* Create few records in this table */
INSERT INTO emp VALUES
(1,100,'Tom'),
(2,200,'Lucy'),
(3,300,'Frank'),
(4,100,'Jane'),
(5,400,'Robert');
我需要得到每个部门的empid、deptid、name和count的结果,如下所示。
我可以使用以下查询获得结果。
SELECT a.empid, a.deptid, a.Name, result.emp_dept_count FROM emp a,
( SELECT b.deptid, COUNT(b.deptid) AS emp_dept_count FROM emp b
GROUP BY b.deptid ) result
WHERE a.deptid = result.deptid;
/* using common table expression */
WITH emp_dept_count_cte(deptid,emp_dept_count) AS ( SELECT b.deptid, COUNT(b.deptid) AS emp_dept_count FROM emp b
GROUP BY b.deptid )
SELECT a.empid, a.deptid, a.Name, result.emp_dept_count
FROM emp a, (SELECT deptid, emp_dept_count FROM emp_dept_count_cte) result
WHERE a.deptid = result.deptid;
/* using common table expression */
WITH emp_dept_count_cte (deptid,emp_dept_count) AS ( SELECT b.deptid, COUNT(b.deptid) AS emp_dept_count FROM emp b
GROUP BY b.deptid )
SELECT a.empid, a.deptid, a.Name, emp_dept_count_cte.emp_dept_count
FROM emp a
INNER JOIN emp_dept_count_cte
ON a.deptid = emp_dept_count_cte.deptid;
/* using common table expression */
WITH emp_dept_count_cte (deptid,emp_dept_count) AS ( SELECT b.deptid, COUNT(b.deptid) AS emp_dept_count FROM emp b
GROUP BY b.deptid )
SELECT a.empid, a.deptid, a.Name, emp_dept_count_cte.emp_dept_count
FROM emp a
LEFT JOIN emp_dept_count_cte
ON a.deptid = emp_dept_count_cte.deptid;
是否有可能以其他方式做到这一点?
【问题讨论】:
-
在您的第一个查询中运行 EXPLAIN EXTENDED [your query],然后是 SHOW WARNINGS;
标签: mysql sql subquery inner-join aggregate-functions