一、基本概念（查询语句）*
①基本语句
“select * from 表名;”，—可查询表中全部数据；
2、“select 字段名 from 表名;”，—可查询表中指定字段的数据；
3、“select distinct 字段名 from 表名;”，—可对表中数据进行去重查询。
4、“select 字段名 from 表名 where 查询条件;”，—可根据条件查询表中指定字段的数据；
②条件查询
1）比较运算符：>, <, >=, <=, =, !=, <>
查询大于18岁的信息
select * from students where age>18;
select id, name,gender from students where age>18;查询小于18岁的信息
select * from students where age<18;查询年龄为18岁的所有学生的名字
select * from students where age=18;
2）逻辑运算符：and, or, not
–18到28之间的学生信息
select * from students where age>18_and age<28:
–18岁以上的女性
select * from students where age>18 and gender="女";
select * from students where age>18 and gender=2;–18以上或者身高查过180(包含)以上
select * from students where age>18 or height>=180;不在18岁以上的女性这个范围内的信息
select * from students where not (age>18 and gender=2);
年龄不是小于或者等于18并且是女性
select * from students where (not age<=18) and gender=2;
3)模糊查询：like， rlike
% 替换1个或者多个
_ 替换1个
查询姓名中 以“小”开始的名字
select name from students where name="小";
select name from students where name like"小%";
查询姓名中有“小”所有的名字
select name from students whece name like "%小%";
查询有2个字的名字
select name from students where name like "__";
查询有3个字的名字
select name from students where name like "__";
查询至少有2个字的名字 select name from
students where name like "__%";
rlike正则
查询以周开始的姓名
select name from students where name rlike "^周.*";
查询以周开始、伦结尾的姓名
select name from students where name rlike "^周.*伦$";
4)范围查询：in，not in，between…and，not between…and
查询年龄为18、34的姓名
select name, age from students where age=18 or age=34;
select name,age from students where age in (18，34);
not in不非连续的范围之内
年龄不是 18、34岁之间的信息
select name,age from students where age not in (18，34);
between … and …表示在一个连续的范围内
查询年龄在18到34之间的的信息
select name,age from students where age between 18 and 34;
not between … and …表示不在一个连续的范围内
查询年龄不在在18到34之间的的信息
select * from students where age not between 18 and 34;
空判断
判空 is null
查询身高为空的信息
select *from students where height is null/NULL/Null;
判非空is not null
select * from students where height is not null;
排序：order_by
–查询年龄在18到34岁之间的男性，按照年龄从小到大排序
select * from students where (age between 18 and 34) and gender=1;
select * from students where (age between 18 and 34) and gender=1 order by age;
select * from students where (age between 18 and 34) and gender=1 order by age asc;
查询年龄在18到34岁之间的女性，身高从高到矮排序
select * from students where (age between18 and 34) and gender=2 order by height desc;
order by多个字段
查询年龄在18到34岁之间的女性，身高从高到矮排序，如果身高相同的情况下按照年龄从小到大排序
select * from students where (age between 18 and 34) and gender=2 order by height desc,age asc;
查询年龄在18到34岁之间的女性，身高从高到矮排序，如果身高相同的情况下按照年龄从小到大排序，如果年龄也相同那么按照id从大到小排序
select * from students where (age between 18 and 34) and gender=2 order by height desc,age asc, id desc;
按照年龄从小到大、身高从高到矮的排序
select * from students order by age asc,height desc;
分组：group_by, group_concat():查询内容， having
where :是对整个数据表信息的判断;
having:是对于分组后的数据进行判断
–group by
按照性别分组,查询所有的性别
select gender from students group by gender;
–计算每种性别中的人数
select gender, count(*) from students group by gender;
where是在group by 前面
–计算男性的人数
select count(*) from students where gender='男';
–group_concat(…)
查询同种性别中的姓名
select gender,group_concat(name) from students group by gender;
having :having是在group by后面
查询平均年龄超过30岁的性别，以及姓名
select gender ,avg(age) from students group by gender having avg(age) > 30;
查询每种性别中的人数多于2个的信息
select gender,count(*) from students group by gender having count(*) > 2;
– 查询每组性别的平均年龄
select gender,avg(age) from students group by gender;
分页： limit
limit start,count (start:表示从哪─个开始;count:表示数量)即limit（第N页-1)*每个的个数，每页的个数;limit在使用的时候，要放在最后面.限制查询出来的数据个数
select *from students where gender=1 limit 2;
查询前5个数据
select* from students limit 0,5;
查询id6-10(包含)的书序
select * from students limit 5,5;
每页显示2个，第1个页面
select * from students limit 0,2;
每页显示2个，第2个页面
select * from students limit 2,2;
每页显示2个，第3个页面
select * from students limit 4,2;
每页显示2个，第4个页面
select * from students limit 6,2;
每页显示2个，显示第6页的信息,按照年龄从小到大排序
select * from students order by age asc limit 10,2;
– 如果重新排序了,那么会显示第一页
select * from students where gender=2 order by height des limit 0,2;
5)聚合函数：count(), max(), min(), sum(), avg(), round()
聚合函数
-总数-- count
-查询男性有多少人，女性有多少人
select count(*) from students where gender=1;
select count(*) as 男性人数 from students where gender=1;
select count(*) as 女性人数 from students where gender=2;
-最大值-最小值
– max --min
一查询最大的年龄
select max (age) from students;
–查询女性的最高身高
select max (height) from students where gender=2;
-求和
–sum
-计算所有人的年龄总和
select sum ( age) from students;
–平均值
– avg
–计算平均年龄
select avg(age) from students;
–计算平均年龄
select sum ( age) / count(* ) from students;
–四舍五入round ( 123.23 ,_1)保留1位小数
–计算所有人的平均年龄,保留2位小数
select round (sum(age)/count(*)，2) from students; select round ( sum(age)/count(*)，3) from students ;
–计算男性的平均身高保留2位小数
select round(avg (height),2) from students where gender=1; select name,round(avg(height)，2) from students where gender=1;
6)连接查询 ：inner join, left join, right join
inner join
select … from 表 A inner join表B;
select * from students inner join classes;
查询有能够对应班级的学生以及班级信息
select * from students inner join classes on students.cls_id=classes.id;
按照要求显示姓名、班级
select students.*, classes.name from students inner join classes on students.cls_id=classes.id;
select students.name，classes.name from students inner join classes on students.cls_id=classes.id;
给数据表起名字
select s.name,c.name from students as s inner join classes as c on s.cls_id=c.id;
查询有能够对应班级的学生以及班级信息，显示学生的所有信息，只显示班级名称
select s.*,c.name from students as s inner join classes as c on s.cls_id=c.id;
在以上的查询中,将班级姓名显示在第1列
select c.name，s.* from students as s inner join classes as c on s.cls_id=c.id;
查询有能够对应班级的学生以及班级信息，按照班级进行排序
select c.xxx s.xxx from student as s inner join clssses as c on … order by …;
select c.name，s.* from students as s inner join classes as c on s.cls_id=c.id order by c.name;
当时同一个班级的时候，按照学生的id进行从小到大排序
select c.name，s.* from students as s inner join classes as c on s.cls_id=c.id order by c.name,s.id;
left join
查询每位学生对应的班级信息
select * from students as s left join classes as c on s.cls_id=c.id;
查询没有对应班级信息的学生
– select … from xxx as s left join xxx as c on… where …
– select … from xxx as s left join xxx as c on… . … having …
select * from students as s left join classes as c on s.cls_id=c.id having c.id is null;
select * from students as s left join classes as c on s.cls_id=c.id where c.id is null;
left join是按照左边的表为基准和右边的表进行查询，查到就显示，查不到就显示为null补充
查询所有字段：select * from 表名; 查询指定字段：select 列1,列2,... from 表名; 使用 as 给字段起别名： select 字段 as 名字.... from 表名; 查询某个表的某个字段：select 表名.字段 .... from表名； 可以通过 as 给表起别名： select 别名.字段 .... from 表名 as 别名; 消除重复行： distinct 字段注意：WHERE子句中是不能用聚集函数作为条件表达式的！二、总结
1、普通查询
（1）命令：select * from <表名>;
（2）命令：select <要查询的字段> from <表名>;
2、去重查询（distinct）
命令：select distinct <要查询的字段> from <表名>
3、排序查询（order by）
升序：asc
降序：desc
降序排列命令：select <要查询的字段名> from <表名> order by <要查询的字段名> desc
不加desc一般默认为升序排列
4、分组查询(group by)
命令：select <按什么分的组>, Sum(score) from <表名> group by <按什么分的组>
假设现在又有一个学生成绩表(result)。要求查询一个学生的总成绩。我们根据学号将他们分为了不同的组。
命令：
select id, Sum(score) from result group by id;现在有两个表学生表（stu）和成绩表（result）。
5.等值查询
当连接运算符为“=”时，为等值连接查询。
现在要查询年龄小于20岁学生的不及格成绩。
select stu.id,score from stu,result where stu.id = result.id and age < 20 and score < 60;等值查询效率太低
6.外连接查询
①语法
select f1,f2,f3,....from table1 left/right outer join table2on 条件;②左外连接查询，例如
select a.id,scorefrom (select id,age from stu where age < 20) a (过滤左表信息)left join (select id, score from result where score < 60) b (过滤右表信息)on a.id = b.id;左外连接就是左表过滤的结果必须全部存在。如果存在左表中过滤出来的数据，右表没有匹配上，这样的话右表就会出现NULL；
③右外连接查询，例如
select a.id,scorefrom (select id,age from stu where age < 20) a (过滤左表信息)right join (select id, score from result where score < 60) b (过滤右表信息)on a.id = b.id;右外连接就是左表过滤的结果必须全部存在
7.内连接查询
①语法
select f1,f2,f3,....from table1 inter join table2on 条件;②例如
select a.id,scorefrom (select id,age from stu where age < 20) a (过滤左表信息)inner join (select id, score from result where score < 60) b (过滤右表信息)on a.id = b.id;8.合并查询
在图书表(t_book)和图书类别表(t_bookType)中
①.union
使用union关键字是，数据库系统会将所有的查询结果合并到一起，然后去掉相同的记录；
select id from t_book union select id from t_bookType;②.union all
使用union all，不会去除掉重复的记录；
select id from t_book union all select id from t_bookType;来源地址：https://blog.csdn.net/weixin_45721753/article/details/127334239
