select count(*) from record where date > '19991201' and date '19991214'and amount > 2000 (25秒) select date,sum(amount) from record group by date (55秒) select count(*) from record where date > '19990901' and place in ('BJ','SH') (27秒)
select count(*) from record where date > '19991201' and date '19991214' and amount > 2000 (14秒) select date,sum(amount) from record group by date (28秒) select count(*) from record where date > '19990901' and place in ('BJ','SH')(14秒)
select count(*) from record where date > '19991201' and date '19991214' and amount > 2000 (26秒) select date,sum(amount) from record group by date (27秒) select count(*) from record where date > '19990901' and place in ('BJ, 'SH')( 1秒)
select count(*) from record where date > '19991201' and date '19991214' and amount > 2000( 1秒) select date,sum(amount) from record group by date (11秒) select count(*) from record where date > '19990901' and place in ('BJ','SH')( 1秒)
select * from record where substring(card_no,1,4)='5378'(13秒) select * from record where amount/30 1000(11秒) select * from record where convert(char(10),date,112)='19991201'(10秒)
select * from record where card_no like '5378%'( 1秒) select * from record where amount 1000*30( 1秒) select * from record where date= '1999/12/01' ( 1秒)
---- 你会发现SQL明显快起来!
---- 2.例:表stuff有200000行,id_no上有非群集索引,请看下面这个SQL:
select count(*) from stuff where id_no in('0','1') (23秒)
---- 分析: ---- where条件中的'in'在逻辑上相当于'or',所以语法分析器会将in ('0','1')转化为id_no ='0' or id_no='1'来执行。我们期望它会根据每个or子句分别查找,再将结果相加,这样可以利用id_no上的索引;但实际上(根据showplan),它却采用了"OR策略",即先取出满足每个or子句的行,存入临时数据库的工作表中,再建立唯一索引以去掉重复行,最后从这个临时表中计算结果。因此,实际过程没有利用id_no上索引,并且完成时间还要受tempdb数据库性能的影响。
select count(*) from stuff where id_no='0' select count(*) from stuff where id_no='1'
---- 得到两个结果,再作一次加法合算。因为每句都使用了索引,执行时间只有3秒,在620000行下,时间也只有4秒。或者,用更好的方法,写一个简单的存储过程: create proc count_stuff as declare @a int declare @b int declare @c int declare @d char(10) begin select @a=count(*) from stuff where id_no='0' select @b=count(*) from stuff where id_no='1' end select @c=@a+@b select @d=convert(char(10),@c) print @d