清空一個表，自增id從1開始
truncate table 表名;

查詢
select 列名 from 表名 where 條件 order by 列名 [desc|asc] limit 跳過條數,查多少條

AS 別名
列名 as 新列名 注意as可以省掉不寫
NULL值查詢
select * from table1 where 字段 is null;
組合列
select concat('No.',id) from stu;
去重復
select distinct 列名 from 表名; 注意：列名，只能跟一個
排序
select * from 表名 order by 列名1 asc,列名2 desc;
加工之后的信息排序
select concat('No.',id) from stu order by concat('No.',id);
常用函數
select concat('My','S','QL'); 連接
select lower('PHP'); 轉小寫
upper() 轉大寫
select replace('www.mysql.com','www','http://www'); 替換
update stu set name = replace( replace(name,'o','0') ,'i' ,'1'); 將name字段所有的o換為0，所有的i換為1

trim()
ltrim()
rtrim()
select abs(-5) 絕對值
select adddate('2013-02-28',1); 2013-03-01
select now();

通配符 "%" 任意 "_"單個
select * from stu where name like '張_';

正則 不支持 \w ...
select * from stu where name regexp '^a[a-z][a-z]$';

區間操作 [not] between ... and ...
select * from stu where age >=18 and age <=19;
select * from stu where age between 18 and 19;
select * from stu where age not between 18 and 19;

[not] in (值1,值2,...)

sum 求和
avg 平均
count(*) 計數
括號中可以是*，也可以列名
如果括號中是列名，則不會統計該列數據為null值的
max 最大值
select max(age) from stu;
min 最小值

*分組查詢
一條語句得到男生和女生分別是多少人?
select sex,count(*) from stu group by sex; #分組后統計
注意：字段只能是用于分組的字段(group by 后的字段) 或聚合函數

按科目分組統計平均成績

分組帶排序
select sex,avg(age) as avg from stu group by sex order by avg desc;

*分組后篩選 having
按性別分組，求平均年齡大于等于6
select sex,avg(age) as avg from stu group by sex having avg>=6;
按性別分組，求平均年齡大于等于6，按平均年齡從大到小排列
select sex,avg(age) as avg from stu group by sex having avg>=6 order by avg desc;

where 與having可以同時出現，where在篩選之前就執行了，having是在分組之后才執行

顯示人數超過15人的班級，按班級人數排序
select grade_id,count(*) from stu group by grade_id having count(*)>15 order by count(*) desc;
select grade_id,count(*) total from stu group by grade_id having total>15 order by total desc;

WHERE -> GROUP BY -> HAVING ->ORDER BY


成績表
create table score (
id int auto_increment primary key,
stu_id int,
subject varchar(50) ,
score int
) engine=myisam default charset=utf8;

INSERT INTO score (stu_id,subject,score) values (1,'php',65),(1,'mysql',95),(1,'linux',80),(2,'php',50),(2,'mysql',70),(2,'linux',70);

統計及格總人數和及格人數的平均分
select subject, count(*),avg(score) from score where score>=70 group bysubject;
統計及格總人數和平均分在80以上


多個學員不低于90分的班級
select 班級 from 表名 where score >=90 group by 班級 having count(*)>1


顯示：id,姓名,班級名稱
select stu.id,stu.name,grade.name from stu,grade where stu.grade_id=grade.id;
顯示：id,姓名,班級名稱,age大于10的
select stu.id,stu.name,grade.name,age from stu,grade where stu.grade_id=grade.id and age>10;

select stu.id,stu.name,grade.name,age from stu
left join grade on grade.id=stu.grade_id
where age>10;

左聯接
左表中的數據，全部出來，就算右表沒，沒有對應的數據
select stu.id,stu.name,grade.name,age from stu
left join grade on grade.id=stu.grade_id

右聯接
select stu.id,stu.name,grade.name,age from stu
right join grade on grade.id=stu.grade_id

用右連接，實現和前面一樣的效果
select stu.id,stu.name,grade.name,age from grade
right join stu on grade.id=stu.grade_id

內聯 inner可以省掉
select stu.id,stu.name,grade.name,age from stu
inner join grade on grade.id=stu.grade_id
內聯與這樣結果相同： select * 表1,表2 where 條件


顯示所有參加考試的男生的成績
select score.*,stu.name from stu join score on score.stu_id=stu.id where sex=1;
select score.*,stu.name from stu ,score where stu.id=score.stu_id and sex=1;
先查男生的id，然后通過id列表，去成績表中查詢
select id from stu where sex=1; select * from score where stu_id in (1,6,8);
select * from score where stu_id in (select id from stu where sex=1);


查詢科目及格人數和不及格人數：
科目 及格人數 不及格人數
phy 20 1
mysql 26 2

select subject,sum(case when score<60 then 1 else 0 end) as no,
sum(case when score>=60 then 1 else 0 end) pass
from score group by subject;

轉載于:https://www.cnblogs.com/wicub/p/4322933.html

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