access数据库中的Round函数的使用方法：
四舍五入的问题
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select round(83.124,2) as t from jqdltb |
显示为:83.12
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1 |
select round(83.125,2) as t from jqdltb |
显示为:83.12
|
1 |
select round(83.126,2) as t from jqdltb |
显示为:83.13
问题来了，就是四舍五入怎么变成五舍六入了呢？
下面是用这个方法来调整达到四舍五入的方法:
使用int函数和加0.005,然后*100，再然后/100
如下所示:
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select int((83.121+0.005)*100)/100 as t from jqdltb |
显示为:83.12
|
1 |
select int((83.124+0.005)*100)/100 as t from jqdltb |
显示为:83.12
|
1 |
select int((83.125+0.005)*100)/100 as t from jqdltb |
显示为:83.13
|
1 |
select int((83.126+0.005)*100)/100 as t from jqdltb |
显示为:83.13
|
1 |
select int((83.129+0.005)*100)/100 as t from jqdltb |
显示为:83.13
