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2207/MySQL数据库操作综合练习题答案.md

@@ -179,6 +179,79 @@ VALUES
 
 ![image-20250315225441802](https://yp.smallkun.cn/markdown/image-20250315225441802.png!compress)
 
+```sql
+#5
+SELECT e.empid, e.`name`, e.title, d.depname, s.basesalary + s.titleSalary, s.basesalary + s.titleSalary-IFNULL(s.decuction, 0)
+FROM employee e, department d, salary s
+WHERE e.depid = d.depid AND e.empid = s.empid;
+#6
+SELECT e.`name`, s.basesalary
+FROM employee e, department d, salary s
+WHERE e.depid = d.depid AND e.empid = s.empid AND d.depname = '销售部';
+#7
+SELECT *
+FROM employee e
+WHERe e.`name` LIKE '张%' AND TIMESTAMPDIFF(YEAR,e.birthday, NOW()) < 40;
+#8
+SELECT e.`name`, s.basesalary, s.titleSalary
+FROM employee e, salary s
+WHERE e.sex = '男' AND e.empid AND s.empid;
+#9
+SELECT e.`name`, e.title, d.depname
+FROM employee e, department d, salary s
+WHERE e.depid = d.depid AND e.empid = s.empid AND s.basesalary < 2000;
+#10
+SELECT COUNT(*)
+FROM employee;
+#11
+SELECT COUNT(*)
+FROM department d;
+#12
+SELECT AVG(s.basesalary + s.titleSalary), MAX(s.basesalary + s.titleSalary), MIN(s.basesalary + s.titleSalary)
+FROM employee e, salary s
+WHERE e.empid = s.empid;
+#13
+SELECT d.depname, AVG(s.basesalary + s.titleSalary)
+FROM employee e, salary s, department d
+WHERE e.empid = s.empid AND e.depid = d.depid
+GROUP BY d.depid;
+#14
+SELECT d.depname, AVG(s.basesalary + s.titleSalary) avg
+FROM employee e, salary s, department d
+WHERE e.empid = s.empid AND e.depid = d.depid
+GROUP BY d.depid
+HAVING avg<2000;
+#15
+SELECT e.empid, e.`name`, s.basesalary, s.titleSalary, s.decuction
+FROM employee e, salary s, department d
+WHERE e.empid = s.empid AND e.depid = d.depid
+ORDER BY s.titleSalary, s.basesalary;
+#16
+SELECT e.empid, e.`name`, e.birthday, 
+	IF(YEAR(e.birthday) < 1980, '老年', 
+	IF(YEAR(e.birthday)< 1990, '中年', '青壮年'))
+FROM employee e;
+#17
+SELECT e.*, depname
+FROM employee e, department d
+WHERE e.depid = d.depid;
+#18
+SELECT d.*, e.*
+FROM employee e, department d
+WHERE e.depid = d.depid;
+#19
+SELECT *
+FROM employee e
+WHERE e.title LIKE '%工程师%' AND e.sex = '男';
+#20
+SELECT d.depname, e.sex, COUNT(*), AVG(s.basesalary)
+FROM employee e, department d, salary s
+WHERE e.depid = d.depid AND e.empid = s.empid
+GROUP BY d.depid, e.sex;
+```
+
+
+
 ---
 
 ### 练习四