查询上面这种表:
1.查询表中所有数据:MariaDB [test]> select * from bug;
2.查询系统为11.3的bug列表:MariaDB [test]> select * from bug where system="iOS11.3";
3.设定返回的记录行数:MariaDB [test]> select * from bug where system="iOS11.3" limit 1;
4.设定开始查询的向下偏移量;MariaDB [test]> SELECT * FROM BUG LIMIT 2 OFFSET 1;
5.模糊查询,查询testname 以3结尾:MariaDB [test]> select * from bug where testername like "%3";
6.从bug表和tester表中,选中不同的name: select testername from bug union select name from tester;
7.从bug表和tester表中,选中重复的name:MariaDB [test]> select testername from bug union all select name from tester;
8.按iD升序排列:MariaDB [test]> select * from bug order by id;
9.按id降序排列:MariaDB [test]> select * from bug order by id desc;
10.按system分组,并统计每种system有多少个:MariaDB [test]> select system,count(*) from bug group by(system);
还可以用sum(),avg()等
11.内连接
select a.runoob_count,b.submission_date from tb1,a inner join tb2,b on a.runoob_author =b.runoob_author;
12.左连接
select a.runoob_count,b.submission_date from tb1,a left join tb2,b on a.runoob_author =b.runoob_author;
13.右连接
select a.runoob_count,b.submission_date from tb1,a right join tb2,b on a.runoob_author =b.runoob_author;