20200324 连接查询作业

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根据课上的emp和dept表完成下列作业:

  1. 查询张姓员工的员工信息和所在部门信息。
  2. 查询张三丰管理了几个员工
  3. 查询出所有实习员工(实习员工无部门信息)

建表语句

DROP TABLE IF EXISTS `emp`;
CREATE TABLE `emp`  (
  `id` int(11) NOT NULL COMMENT '员工编号',
  `name` varchar(255) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL COMMENT '员工姓名',
  `dept_id` int(11) NULL DEFAULT NULL COMMENT '部门编号',
  `leader` int(11) NULL DEFAULT NULL COMMENT '直属领导id',
  `is_enable` int(11) NULL DEFAULT NULL COMMENT '是否在职 1在职 0离职',
  PRIMARY KEY (`id`) USING BTREE
) ENGINE = InnoDB CHARACTER SET = utf8 COLLATE = utf8_general_ci ROW_FORMAT = Compact;


INSERT INTO `emp` VALUES (1, '张三丰', 1, 0, 1);
INSERT INTO `emp` VALUES (2, '张无忌', 1, 1, 1);
INSERT INTO `emp` VALUES (3, '小龙女', 1, 1, 1);
INSERT INTO `emp` VALUES (4, '小白菜', 1, 3, 1);
INSERT INTO `emp` VALUES (5, '韦小宝', 2, 0, 1);
INSERT INTO `emp` VALUES (6, '令狐冲', 2, 0, 1);
INSERT INTO `emp` VALUES (7, '东方不败', 0, 8, 1);
INSERT INTO `emp` VALUES (8, '任我行', 3, 0, 1);
INSERT INTO `emp` VALUES (9, '李寻欢', 0, 8, 1);


DROP TABLE IF EXISTS `dept`;
CREATE TABLE `dept`  (
  `id` int(11) NOT NULL COMMENT '部门id',
  `name` varchar(255) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL COMMENT '部门名称',
  PRIMARY KEY (`id`) USING BTREE
) ENGINE = InnoDB CHARACTER SET = utf8 COLLATE = utf8_general_ci ROW_FORMAT = Compact;

INSERT INTO `dept` VALUES (1, '销售部');
INSERT INTO `dept` VALUES (2, '信息技术部');
INSERT INTO `dept` VALUES (3, '财务部');
INSERT INTO `dept` VALUES (4, '有关部门');

上课实战脚本

#交叉连接
select * from emp;
#笛卡尔积
select * from emp,dept;
select * from emp cross join dept;
#内连接
#显示内连接,标准内连接 on 两张表如何关联
select * from emp as a inner join dept as b on a.dept_id = b.id;
#隐式内连接 where 在结果集的基础上做条件筛选
select * from emp as a,dept as b where a.dept_id = b.id;
#内连接 join 省略 inner 
select * from emp as a join dept as b on a.dept_id = b.id;
#内连接 cross join
select * from emp as a cross join dept as b on a.dept_id = b.id;
# mysql cross join on  inner join on 结果上没有区别
# 标准sql cross join 不能使用onmysql中支持on的操作。
#特殊内连接 不等值内连接
select * from emp as a inner join dept as b on a.dept_id > b.id;
#特殊内连接 自连接
select * from emp as a inner join emp as b on a.id = b.leader;
#外连接
#外连接显示的内容比内连接要多,内连接的补充 >=
#左外连接,左表为主表右表为从表,主表所有数据都显示,从表只有匹配上的数据才显示
select * from emp as a left join dept as b on a.dept_id = b.id;
#左外连接,右表为主表左表为从表,主表所有数据都显示,从表只有匹配上的数据才显示
select * from emp as a right join dept as b on a.dept_id = b.id;
select * from dept as b left join emp as a  on a.dept_id = b.id;
#外连接 查询从表为空的数据
select * from emp as a left join dept as b on a.dept_id = b.id
where b.id is null;
# on(两张表如何关联)onwhere之前  where(结果集的基础上做条件筛选)的区别 
select * from emp as a left join dept as b 
on a.dept_id = b.id where b.id is null;
#全连接 full join 
#sql1
#UNION
#sql2;
#sql1 + sql2 
select * from emp as a left join dept as b on a.dept_id = b.id
union
select * from emp as a right join dept as b on a.dept_id = b.id;
# union all 不去重复
select * from emp as a left join dept as b on a.dept_id = b.id
union all
select * from emp as a right join dept as b on a.dept_id = b.id;
1 操作
luojie 在 2020-08-06 17:33:22 更新了该帖
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  • luojie

    满分~

  • 其他回帖
  • cherrybing

    -- 1、查询张姓员工的员工信息和所在部门信息

    SELECT e.name,e.leader,e.is_enable,dt.nameas 部门 from emp e
    left JOIN dept dt on e.dept_id=dt.id
    where e.name like '张%';

    -- 查询张三丰管理了几个员工

    #方法一:

    SELECT count(*) from emp ep where ep.leader in(SELECT id from emp where name='张三丰');

    #方法二:

    SELECT count(em1.name) as 所管理员工数,em2.name as 领导 from emp em1
    LEFT JOIN emp em2 on em1.leader=em2.id
    GROUP BY em2.name
    having em2.name='张三丰';

    -- 查询出所有实习员工(实习员工无部门信息)

    SELECT ep.*,dt.nameas 部门名称 from emp ep
    LEFT JOIN dept dt on ep.dept_id=dt.id
    where dt.name is NULL

  • stern

    image.png
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  • 13207145280

    第一题:select * from emp as a inner join dept as
    b on a.dept_id=b.id where a.name like '张%';
    image.png

    第二题:select * from emp as a where dept_id=1 and leader=1;
    image.png

    不知道怎么计数

    第三题:select * from emp as a where dept_id=1 and leader=1
    image.png

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