Question

mysql 嵌套查询

select
train_no,a.sequ_no from,b.sequ_no to
from
(select * from aa where st_name= "xxx" ) as a ,
(select * from aa where st_name= "xx") as b
where
a.train_no =b.train_no and a.sequ_no<b.sequ_no;
当我这样写时，报错，请问是什么原因呢？ 错误位置始终在 第二行到 第四行 这里

Answer 1

你的sql语句中 的 from 和 to 与sql保留关键词 有冲突。

select
a.train_no, a.sequ_no t_from , b.sequ_no t_to
from
(select * from aa where st_name= "xxx" ) as a ,
(select * from aa where st_name= "xx") as b
where
a.train_no = b.train_no and a.sequ_no<b.sequ_no;

或者

select
a.train_no, a.sequ_no as `from` , b.sequ_no as `to`
from
(select * from aa where st_name= "xxx" ) a inner join
(select * from aa where st_name= "xx") b on a.train_no =b.train_no
where a.sequ_no<b.sequ_no;