mysql(单表查询,多表查询,MySQl创建用户和授权,可视化工具Navicat的使用)

时间:2023-03-09 07:52:04
mysql(单表查询,多表查询,MySQl创建用户和授权,可视化工具Navicat的使用)

单表查询

语法:

一、单表查询的语法
SELECT 字段1,字段2... FROM 表名
WHERE 条件
GROUP BY field
HAVING 筛选
ORDER BY field
LIMIT 限制条数
二、关键字的执行优先级(重点) 重点中的重点:关键字的执行优先级
from
where
group by
having
select
distinct
order by
limit

1.找到表:from

2.拿着where指定的约束条件,去文件/表中取出一条条记录

3.将取出的一条条记录进行分组group by,如果没有group by,则整体作为一组

4.将分组的结果进行having过滤

5.执行select

6.去重

7.将结果按条件排序:order by

8.限制结果的显示条数

创建公司员工表,表的字段和数据类型

company.employee
员工id id int
姓名 name varchar
性别 sex enum
年龄 age int
入职日期 hire_date date
岗位 post varchar
职位描述 post_comment     varchar
薪水 salary     double
办公室 office int
部门编号 depart_id int 

创建员工表,并插入记录

#创建表,设置字段的约束条件
create table employee(
id int primary key auto_increment,
name varchar(20) not null,
sex enum('male','female') not null default 'male', #大部分是男的
age int(3) unsigned not null default 28,
hire_date date not null,
post varchar(50),
post_comment varchar(100),
salary double(15,2),
office int,#一个部门一个屋
depart_id int
);
# 查看表结构
mysql> desc employee;
+--------------+-----------------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+--------------+-----------------------+------+-----+---------+----------------+
| id | int(11) | NO | PRI | NULL | auto_increment |
| emp_name | varchar(20) | NO | | NULL | |
| sex | enum('male','female') | NO | | male | |
| age | int(3) unsigned | NO | | 28 | |
| hire_date | date | NO | | NULL | |
| post | varchar(50) | YES | | NULL | |
| post_comment | varchar(100) | YES | | NULL | |
| salart | double(15,2) | YES | | NULL | |
| office | int(11) | YES | | NULL | |
| depart_id | int(11) | YES | | NULL | |
+--------------+-----------------------+------+-----+---------+----------------+
rows in set (0.08 sec) #插入记录
#三个部门:教学,销售,运营
insert into employee(name ,sex,age,hire_date,post,salary,office,depart_id) values
('aa','male',18,'','财务',7300.33,401,1),
('bb','male',78,'','teacher',1000000.31,401,1),
('cc','male',81,'','teacher',8300,401,1),
('dd','male',73,'','teacher',3500,401,1),
('ee','male',28,'','teacher',2100,401,1),
('ff','female',18,'','teacher',9000,401,1),
('gg','male',18,'','teacher',30000,401,1),
('hh','male',48,'','teacher',10000,401,1), ('AA','female',48,'','sale',3000.13,402,2),#以下是销售部门
('BB','female',38,'','sale',2000.35,402,2),
('CC','female',18,'','sale',1000.37,402,2),
('DD','female',18,'','sale',3000.29,402,2),
('EE','female',28,'','sale',4000.33,402,2), ('FF','male',28,'','operation',10000.13,403,3), #以下是运营部门
('GG','male',18,'','operation',20000,403,3),
('HH','female',18,'','operation',19000,403,3),
('II','male',18,'','operation',18000,403,3),
('JJ','female',18,'','operation',17000,403,3)
; 创建员工表,并插入记录

(1)where 约束 

where子句中可以使用
1.比较运算符:>、<、>=、<=、<>、!=
2.between 80 and 100 :值在80到100之间
3.in(80,90,100)值是10或20或30
4.like 'xiaomagepattern': pattern可以是%或者_。%小时任意多字符,_表示一个字符
5.逻辑运算符:在多个条件直接可以使用逻辑运算符 and or not

验证结果:

where约束

#1 :单条件查询
mysql> select id,name from employee where id > 5; #2 多条件查询
mysql> select name from employee where post='teacher' and salary>10000; #3.关键字BETWEEN AND
SELECT name,salary FROM employee
WHERE salary BETWEEN 10000 AND 20000; SELECT name,salary FROM employee
WHERE salary NOT BETWEEN 10000 AND 20000; #注意''是空字符串,不是null
SELECT name,post_comment FROM employee WHERE post_comment='';
ps:
执行
update employee set post_comment='' where id=2;
再用上条查看,就会有结果了
#5:关键字IN集合查询
mysql> SELECT name,salary FROM employee WHERE salary=3000 OR salary=3500 OR salary=4000 OR salary=9000 ; mysql> SELECT name,salary FROM employee WHERE salary IN (3000,3500,4000,9000) ;
mysql> SELECT name,salary FROM employee WHERE salary NOT IN (3000,3500,4000,9000) ;#6:关键字LIKE模糊查询
通配符’%’ 可以匹配任意多字符
mysql> SELECT * FROM employee WHERE name LIKE 'a%'; 通配符'_' 只能匹配一个字符
 mysql> SELECT age FROM employee WHERE name LIKE 'ale_'; +-----+ | age | +-----+ | 78 | +-----+ row in set (0.00 sec) 
练习:
1. 查看岗位是teacher的员工姓名、年龄 2. 查看岗位是teacher且年龄大于30岁的员工姓名、年龄 3. 查看岗位是teacher且薪资在9000-1000范围内的员工姓名、年龄、薪资 4. 查看岗位描述不为NULL的员工信息 5. 查看岗位是teacher且薪资是10000或9000或30000的员工姓名、年龄、薪资 6. 查看岗位是teacher且薪资不是10000或9000或30000的员工姓名、年龄、薪资 7. 查看岗位是teacher且名字是jin开头的员工姓名、年薪 #对应的sql语句
select name,age from employee where post = 'teacher';
select name,age from employee where post='teacher' and age > 30;
select name,age,salary from employee where post='teacher' and salary between 9000 and 10000;
select * from employee where post_comment is not null;
select name,age,salary from employee where post='teacher' and salary in (10000,9000,30000);
select name,age,salary from employee where post='teacher' and salary not in (10000,9000,30000);
select name,salary*12 from employee where post='teacher' and name like 'jin%'; where约束

(2)group by 分组查询

#1、首先明确一点:分组发生在where之后,即分组是基于where之后得到的记录而进行的
#2、分组指的是:将所有记录按照某个相同字段进行归类,比如针对员工信息表的职位分组,或者按照性别进行分组等
#3、为何要分组呢?
取每个部门的最高工资
取每个部门的员工数
取男人数和女人数
小窍门:‘每’这个字后面的字段,就是我们分组的依据
#4、大前提:
可以按照任意字段分组,但是分组完毕后,比如group by post,只能查看post字段,如果想查看组内信息,需要借助于聚合函数

当执行以下sql语句的时候,是以post字段查询了组中的第一条数据,没有任何意义,因为我们现在想查出当前组的多条记录。

mysql> select * from employee group by post;#由于没有设置ONLY_FULL_GROUP_BY,于是也可以有结果,默认都是组内的第一条记录,但其实这是没有意义的
如果想分组,则必须要设置全局的sql的模式为ONLY_FULL_GROUP_BY
mysql> set global sql_mode='ONLY_FULL_GROUP_BY';
Query OK, 0 rows affected (0.00 sec) #查看MySQL 5.7默认的sql_mode如下:
mysql> select @@global.sql_mode;
+--------------------+
| @@global.sql_mode |
+--------------------+
| ONLY_FULL_GROUP_BY |
+--------------------+
1 row in set (0.00 sec) mysql> exit;#设置成功后,一定要退出,然后重新登录方可生效
Bye

继续验证通过group by分组之后,只能查看当前字段,如果想查看组内信息,需要借助于聚合函数

mysql> select * from emp group by post;# 报错
ERROR 1054 (42S22): Unknown column 'post' in 'group statement' mysql> select post from employee group by post;
+-----------------------------------------+
| post |
+-----------------------------------------+
| operation |
| sale |
| teacher |
| 财务 |
+-----------------------------------------+
4 rows in set (0.00 sec)

(3)聚合函数

max()求最大值
min()求最小值
avg()求平均值
sum() 求和
count() 求总个数
group_concat(name) 查询名字 #强调:聚合函数聚合的是组的内容,若是没有分组,则默认一组
# 每个部门有多少个员工以及员工的名字
select post,count(id),group_concat(name) from employee group by post;
select post,count(1) from employee group by post; #加了count(1)后,会在employee表的末尾添加一列1,数1的效果和数id是一样的。
# 每个部门的最高薪水
select post,max(salary) from employee group by post;
# 每个部门的最低薪水
select post,min(salary) from employee group by post;
# 每个部门的平均薪水
select post,avg(salary) from employee group by post;
# 每个部门的所有薪水
select post,sum(age) from employee group by post;

小练习

1. 查询岗位名以及岗位包含的所有员工名字
  select post,name from employee;
2. 查询岗位名以及各岗位内包含的员工个数
   select post ,count(1) from employee group by post;
3. 查询公司内男员工和女员工的个数
  select sex,count(1) from employee group by sex;
4. 查询岗位名以及各岗位的平均薪资
  select post,avg(salary) from employee group by post;
5. 查询岗位名以及各岗位的最高薪资
  select post,max(salary) from employee group by post;
6. 查询岗位名以及各岗位的最低薪资
   select post,min(salary) from employee group by post;
7. 查询男员工与男员工的平均薪资,女员工与女员工的平均薪资
  select sex,avg(salary) from employee group by sex;

(4)HAVING过滤

HAVING与WHERE不一样的地方在于

#!!!执行优先级从高到低:where > group by > having
#1. Where 发生在分组group by之前,因而Where中可以有任意字段,但是绝对不能使用聚合函数。 #2. Having发生在分组group by之后,因而Having中可以使用分组的字段,无法直接取到其他字段,可以使用聚合函数

验证:

验证:
mysql> select * from employee where salary>1000000;
+----+------+------+-----+------------+---------+--------------+------------+--------+-----------+
| id | name | sex | age | hire_date | post | post_comment | salary | office | depart_id |
+----+------+------+-----+------------+---------+--------------+------------+--------+-----------+
| 2 | bb | male | 78 | 2015-03-02 | teacher | | 1000000.31 | 401 | 1 |
+----+------+------+-----+------------+---------+--------------+------------+--------+-----------+
row in set (0.00 sec) mysql> select * from employee having salary>1000000; # 现在不报错
ERROR 1463 (42000): Non-grouping field 'salary' is used in HAVING clause # 必须使用group by才能使用group_concat()函数,将所有的name值连接
mysql> select post,group_concat(name) from employee group by post having salary > 10000; ##错误,分组后无法直接取到salary字段
ERROR 1054 (42S22): Unknown column 'post' in 'field list'
select post,group_concat(name) from employee group by post ; #正确

小练习:

1. 查询各岗位内包含的员工个数小于2的岗位名、岗位内包含员工名字、个数
2. 查询各岗位平均薪资大于10000的岗位名、平均工资
3. 查询各岗位平均薪资大于10000且小于20000的岗位名、平均工资

小练习答案:

# 题1:
mysql> select post,group_concat(name),count(id) from employee group by post; # 造了一个虚拟表,关联了post和name mysql> select post,group_concat(name),count(id) from employee group by post having count(id)<2; #答案这一行就完事了。重在理解。#题2:
mysql> select post,avg(salary) from employee group by post having avg(salary) > 10000;
    select post ,avg(salary) as A from employee group by post having A>10000;
+-----------+---------------+
| post | avg(salary) |
+-----------+---------------+
| operation | 16800.026000 |
| teacher | 151842.901429 |
+-----------+---------------+
rows in set (0.00 sec) #题3:
mysql> select post,avg(salary) from employee group by post having avg(salary) > 10000 and avg(salary) <20000;
     select post ,avg(salary) as A from employee group by post having A between 10000 and 20000;
+-----------+--------------+
| post | avg(salary) |
+-----------+--------------+
| operation | 16800.026000 |
+-----------+--------------+
row in set (0.00 sec)

(5)order by 查询排序

按单列排序
SELECT * FROM employee ORDER BY age;
SELECT * FROM employee ORDER BY age ASC;
SELECT * FROM employee ORDER BY age DESC;
按多列排序:先按照age升序排序,如果年纪相同,则按照id降序
SELECT * from employee
ORDER BY age ASC,
id DESC;

验证多加序列

验证多列排序:
SELECT * from employee ORDER BY age ASC,id DESC;
mysql> SELECT * from employee ORDER BY age ASC,id DESC;

小练习:

1. 查询所有员工信息,先按照age升序排序,如果age相同则按照hire_date降序排序
  select * from employee order by age asc,hire_date desc;
2. 查询各岗位平均薪资大于10000的岗位名、平均工资,结果按平均薪资升序排列
  select post,avg(salary) as agv_ from employee group by post having agv_ >10000 order by agv_  asc;
3. 查询各岗位平均薪资大于10000的岗位名、平均工资,结果按平均薪资降序排列 

小练习答案

# 题目1
select * from employee ORDER BY age asc,hire_date desc; # 题目2
mysql> select post,avg(salary) from employee group by post having avg(salary) > 10000 order by avg(salary) asc; #关键字优先级要熟记
    
+-----------+---------------+
| post | avg(salary) |
+-----------+---------------+
| operation | 16800.026000 |
| teacher | 151842.901429 |
+-----------+---------------+
rows in set (0.00 sec) # 题目3
mysql> select post,avg(salary) from employee group by post having avg(salary) > 10000 order by avg(salary) desc;
+-----------+---------------+
| post | avg(salary) |
+-----------+---------------+
| teacher | 151842.901429 |
| operation | 16800.026000 |
+-----------+---------------+
rows in set (0.00 sec) mysql>

(5)limit  限制查询的记录数:

示例:
SELECT * FROM employee ORDER BY salary DESC
LIMIT 3; #默认初始位置为0 SELECT * FROM employee ORDER BY salary DESC
LIMIT 0,5; #从第0开始,即先查询出第一条,然后包含这一条在内往后查5条 SELECT * FROM employee ORDER BY salary DESC
LIMIT 5,5; #从第5开始,即先查询出第6条,然后包含这一条在内往后查5条

小练习:
分页显示,每页5条

小练习答案

# 第1页数据
mysql> select * from employee limit 0,5;# 第2页数据
mysql> select * from employee limit 5,5;# 第3页数据
mysql> select * from employee limit 10,5;
+----+-----------+--------+-----+------------+-----------+--------------+----------+--------+-----------+ 小练习答案

多表查询

  • 多表连接查询
  • 符合条件连接查询
  • 子查询

准备工作:准备两张表,部门表(department)、员工表(employee)

create table department(
id int,
name varchar(20)
); create table employee(
id int primary key auto_increment,
name varchar(20),
sex enum('male','female') not null default 'male',
age int,
dep_id int
); #插入数据
insert into department values
(200,'技术'),
(201,'人力资源'),
(202,'销售'),
(203,'运营'); insert into employee(name,sex,age,dep_id) values
('AA','male',18,200),
('BB','female',48,201),
('CC','male',38,201),
('DD','female',28,202),
('EE','male',18,200),
('FF','female',18,204)
; # 查看表结构和数据
mysql> desc department;
+-------+-------------+------+-----+---------+-------+
| Field | Type | Null | Key | Default | Extra |
+-------+-------------+------+-----+---------+-------+
| id | int(11) | YES | | NULL | |
| name | varchar(20) | YES | | NULL | |
+-------+-------------+------+-----+---------+-------+
2 rows in set (0.19 sec) mysql> desc employee;
+--------+-----------------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+--------+-----------------------+------+-----+---------+----------------+
| id | int(11) | NO | PRI | NULL | auto_increment |
| name | varchar(20) | YES | | NULL | |
| sex | enum('male','female') | NO | | male | |
| age | int(11) | YES | | NULL | |
| dep_id | int(11) | YES | | NULL | |
+--------+-----------------------+------+-----+---------+----------------+
5 rows in set (0.01 sec) mysql> select * from department;
+------+--------------+
| id | name |
+------+--------------+
| 200 | 技术 |
| 201 | 人力资源 |
| 202 | 销售 |
| 203 | 运营 |
+------+--------------+
4 rows in set (0.02 sec) mysql> select * from employee;

ps:观察两张表,发现department表中id=203部门在employee中没有对应的员工,发现employee中id=6的员工在department表中没有对应关系。

一、多表连接查询

两张表的准备工作已完成,比如现在我要查询的员工信息以及该员工所在的部门。从该题中,我们看出既要查员工又要查该员工的部门,肯定要将两张表进行连接查询,多表连接查询。

重点:外链接语法

语法:

SELECT 字段列表
FROM 表1 INNER|LEFT|RIGHT JOIN 表2
ON 表1.字段 = 表2.字段;

(1)先看第一种情况交叉连接:不适用任何匹配条件。生成笛卡尔积(关于笛卡尔积的含义,大家百度自行补脑)。

mysql> select * from employee,department;  # 不符合实际,需要刷选

(2)内连接:只连接匹配的行

#找两张表共有的部分,相当于利用条件从笛卡尔积结果中筛选出了匹配的结果
#department没有204这个部门,因而employee表中关于204这条员工信息没有匹配出来
mysql> select employee.id,employee.name,employee.age,employee.sex,department.name from employee inner join department
on employee.dep_id=department.id;#上述sql等同于
mysql> select employee.id,employee.name,employee.age,employee.sex,department.name from employee,department where
employee.dep_id=department.id;

(3)外链接之左连接:优先显示左表全部记录

#以左表为准,即找出所有员工信息,当然包括没有部门的员工
#本质就是:在内连接的基础上增加左边有,右边没有的结果
mysql> select employee.id,employee.name,department.name as depart_name from employee left join department on employee.dep_id=department.id;
6 rows in set (0.00 sec)

(4) 外链接之右连接:优先显示右表全部记录

#以右表为准,即找出所有部门信息,包括没有员工的部门
#本质就是:在内连接的基础上增加右边有,左边没有的结果
mysql> select employee.id,employee.name,department.name as depart_name from employee right join department on employee.dep_id=department.id;

(5) 全外连接:显示左右两个表全部记录(了解)

#外连接:在内连接的基础上增加左边有右边没有的和右边有左边没有的结果
#注意:mysql不支持全外连接 full JOIN
#强调:mysql可以使用此种方式间接实现全外连接
语法:select * from employee left join department on employee.dep_id = department.id
union all
select * from employee right join department on employee.dep_id = department.id; mysql> select * from employee left join department on employee.dep_id = department.id
union
select * from employee right join department on employee.dep_id = department.id
;#注意 union与union all的区别:union会去掉相同的纪录

二、符合条件连接查询

示例1:以内连接的方式查询employee和department表,并且employee表中的age字段值必须大于25,即找出年龄大于25岁的员工以及员工所在的部门

 select employee.name,department.name from employee,department 
    where employee.dep_id=department.id and age>25;

示例2:以内连接的方式查询employee和department表,并且以age字段的升序方式显示。

select employee.id,employee.name,employee.age,department.name from employee,department
where employee.dep_id = department.id
and age > 25
order by age asc;

三、子查询

#1:子查询是将一个查询语句嵌套在另一个查询语句中。
#2:内层查询语句的查询结果,可以为外层查询语句提供查询条件。
#3:子查询中可以包含:IN、NOT IN、ANY、ALL、EXISTS 和 NOT EXISTS等关键字
#4:还可以包含比较运算符:= 、 !=、> 、<等

例子:

(1)带in关键字的子查询

#查询平均年龄在25岁以上的部门名
select id,name from department
where id in
(select dep_id from employee group by dep_id having avg(age) > 25);
# 查看技术部员工姓名
select name from employee
where dep_id in
(select id from department where name='技术');
select dep_id,name from employee
where dep_id in
(select id from department group by id having id=200);
#查看不足1人的部门名 这道题正面不好解答,用逆向思维会容易点。
  select name from department where id not in(select dep_id from employee group by dep_id having count(1)>0);

(2)带比较运算符的子查询

#比较运算符:=、!=、>、>=、<、<=、<>
#查询大于所有人平均年龄的员工名与年龄
mysql> select name,age from employee where age > (select avg(age) from employee);
# 不分组也可以用聚合函数(select avg(age) from employee)
#查询大于部门内平均年龄的员工名、年龄 思路: 
(1)先对员工表(employee)中的人员分组(group by),查询出dep_id以及平均年龄。
(2)将查出的结果作为临时表,再对根据临时表的dep_id和employee的dep_id作为筛选条件将employee表和临时表进行内连接。
(3)最后再将employee员工的年龄是大于平均年龄的员工名字和年龄筛选。
mysql> select t1.name,t1.age from employee as t1
      inner join (select dep_id,avg(age) as avg_age from employee group by dep_id) as t2
      on t1.dep_id = t2.dep_id
      where t1.age > t2.avg_age;

(3)带EXISTS关键字的子查询

#EXISTS关字键字表示存在。在使用EXISTS关键字时,内层查询语句不返回查询的记录。而是返回一个真假值。True或False
#当返回True时,外层查询语句将进行查询;当返回值为False时,外层查询语句不进行查询
#department表中存在dept_id=203,Ture
mysql> select * from employee where exists (select id from department where id=200);
#department表中存在dept_id=205,False
mysql> select * from employee where exists (select id from department where id=204);
Empty set (0.00 sec)

小练习:

查询每个部门最新入职的那位员工

#创建表
create table employee(
id int not null unique auto_increment,
name varchar(20) not null,
sex enum('male','female') not null default 'male', #大部分是男的
age int(3) unsigned not null default 28,
hire_date date not null,
post varchar(50),
post_comment varchar(100),
salary double(15,2),
office int, #一个部门一个屋子
depart_id int
); #查看表结构
mysql> desc employee;
+--------------+-----------------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+--------------+-----------------------+------+-----+---------+----------------+
| id | int(11) | NO | PRI | NULL | auto_increment |
| name | varchar(20) | NO | | NULL | |
| sex | enum('male','female') | NO | | male | |
| age | int(3) unsigned | NO | | 28 | |
| hire_date | date | NO | | NULL | |
| post | varchar(50) | YES | | NULL | |
| post_comment | varchar(100) | YES | | NULL | |
| salary | double(15,2) | YES | | NULL | |
| office | int(11) | YES | | NULL | |
| depart_id | int(11) | YES | | NULL | |
+--------------+-----------------------+------+-----+---------+----------------+

小练习答案

select * from employee as t1
inner join
(select post,max(hire_date) as new_date from employee group by post) as t2
on t1.post=t2.post
where t1.hire_date=t2.new_date;

MySQL创建用户和授权

权限管理

select * from mysql.user\G;  #结构化展示账户信息

我们知道我们的最高权限管理者是root用户,它拥有着最高的权限操作。包括select、update、delete、update、grant等操作。那么一般情况在公司之后DBA工程师会创建一个用户和密码,让你去连接数据库的操作,并给当前的用户设置某个操作的权限(或者所有权限)。那么这时就需要我们来简单了解一下:

  • 如何创建用户和密码
  • 给当前的用户授权
  • 移除当前用户的权限

如果你想创建一个新的用户,则需要以下操作:

1.进入到mysql数据库下

mysql> use mysql
Database changed

2.对新用户增删改

1.创建用户(本地):
create user 'xxx'@'localhost' identified by '123';  #创建本地用户
 创建用户(远程用户)
# 指定ip:192.118.1.1的mjj用户登录 
create user 'xk'@'192.118.1.1' identified by ''; # 指定ip:192.118.1.开头的mjj用户登录 c
reate user 'xk'@'192.118.1.%' identified by '';
# 指定任何ip的mjj用户登录 create user 'xk'@'%' identified by ''; 2.删除用户 drop user '用户名'@'IP地址';
3.修改用户 rename user '用户名'@'IP地址' to '新用户名'@'IP地址';
4.修改密码 set password for '用户名'@'IP地址'=Password('新密码');

3.对当前的用户授权管理

#查看权限
show grants for '用户'@'IP地址'

+-----------------------------------------+
| Grants for xk@localhost |
+-----------------------------------------+
| GRANT USAGE ON *.* TO 'ooo'@'localhost' |
+-----------------------------------------

#授权 mjj用户仅对db1.t1文件有查询、插入和更新的操作
grant all privileges  on *.*  to "xxx"@'localhost';  #本地用户授权
 grant select ,insert,update on db1.t1 to "xk"@'%';
# 表示有所有的权限,除了grant这个命令,这个命令是root才有的。mjj用户对db1下的t1文件有任意操作
grant all privileges on db1.t1 to "xk"@'%';
#mjj用户对db1数据库中的文件执行任何操作
grant all privileges on db1.* to "xk"@'%';
#mjj用户对所有数据库中文件有任何操作
grant all privileges on *.* to "xk"@'%'; #取消权限 # 取消mjj用户对db1的t1文件的任意操作
revoke all on db1.t1 from 'xk'@"%"; # 取消来自远程服务器的mjj用户对数据库db1的所有表的所有权限 revoke all on db1.* from 'xk'@"%"; 取消来自远程服务器的mjj用户所有数据库的所有的表的权限
revoke all privileges on *.* from 'xk'@'%';
授权(取消权限)后一定要刷新权限
FLUSH PRIVILEGES;

ps:在公司中,一般情况下是DBA工程师来做这些授权工作。给你一个用户名和密码,你来连接就可以了。

4.MySql备份命令行操作

# 备份:数据表结构+数据
mysqdump -u root db1 > db1.sql -p
# 备份:数据表结构
mysqdump -u root -d db1 > db1.sql -p #导入现有的数据到某个数据库
#1.先创建一个新的数据库
create database db10;
# 2.将已有的数据库文件导入到db10数据库中
mysql -u root -d db10 < db1.sql -p

可视化工具Navicat的使用

  • 掌握Navicat的基本使用
# PS:在生产环境中操作MySQL数据库还是推荐使用命令行工具mysql,但在我们自己开发测试时,可以使用可视化工具Navicat,以图形界面的形式操作MySQL数据库 

官网下载:https://www.navicat.com/en/products/navicat-for-mysql

需要掌握基本的操作

掌握:
#1. 测试+链接数据库
#2. 新建库
#3. 新建表,新增字段+类型+约束
#4. 设计表:外键
#5. 新建查询
#6. 备份库/表 #注意:
批量加注释:ctrl+?键
批量去注释:ctrl+shift+?键
分类: MySql