大家有什么高效率的实现办法吗?
我只知道
select top 500 from table order by id ase
update table set isuse = 1 where id < (select max(id) from (select top 500 from table order by id ase))
思路是这样的,大家给点意见
我只知道
select top 500 from table order by id ase
update table set isuse = 1 where id < (select max(id) from (select top 500 from table order by id ase))
思路是这样的,大家给点意见
update [table] set isuse = 1
where id in(select top 500 id from [table] order by id asc)
update [table] set isuse = 1
where id < (select max(id) from (select top 500 id from [table] order by id )t)
update [table] set isuse = 1
where id <=(select max(id) from (select top 500 id from [table] order by id )t)
lsqkeke(可可)
有没有考虑过两种方法哪个效率会高一些呢?
----------------------
你的方法高效一点!
因为:SQL在执行计划中:
select max(id) from (select top 500 id from [table] order by id )t
该语句总共执行一次! 作大小判断时,也是一次!而我的写法呢,在id 是否包含在后面的集合时,花了较多的时间