sql - 获取从该日期到该日期的行

Question

我真的无法理解。

代码-

select * 
from Bill_Detail 
where DateTimeofBilling > '8/18/2013 12:00:00 PM' 
  and DateTimeofBilling < '8/20/2013 12:00:00 AM'

伪代码-

select all 
from Bill_Detail 
where DateTimeofBilling greater than 8/18/2013 
  and DateTimeofBilling less than 8/20/2013

我需要从日期 8/18/2013 到 8/20/2013 获取行，但代码没有返回任何内容。

我认为这足以解释这一点。谁能帮忙？

Answer 1

为什么您使用PM（中午，而不是午夜）进行“大于”检查？您的伪代码假定您还想使用 PM。

select * 
from Bill_Detail 
where DateTimeofBilling >= '20130818'-- if you want to include this day completely
  and DateTimeofBilling < '20130820' -- if you want to exclude this day completely

Answer 2

尝试这个..

select * from Bill_Detail where DateTimeofBilling > '2013/08/18 12:00:00 PM' and DateTimeofBilling < '2013/08/20 12:00:00 AM'

Answer 3

如果您的列DateTimeofBilling是 datetime 类型，请将其与 date 而不是 varchar 进行比较：

select * 
from Bill_Detail 
where
    DateTimeofBilling >= convert(date, '20130818', 112) and
    DateTimeofBilling < convert(date, '20130821', 112)

这将返回带有 DateTimeofBilling 的行，其中日期大于或等于2013-08-18且小于或等于2013-08-20

Answer 4

之间使用

SELECT *
FROM Bill_Detail
WHERE DateTimeofBilling BETWEEN '20130818 000000' 
    AND DateTimeofBilling < '2013/08/20 12:00:00 AM'

Answer 5

尝试这个，

select * from Bill_Detail where DateTimeofBilling between '8/18/2013 12:00:00 PM' and  '8/20/2013 12:00:00 AM'

Answer 6

SELECT * FROM Bill_Detail WHERE DateTimeofBilling>='8/18/2013 12:00:00 PM' AND DateTimeofBilling<='8/20/2013 12:00:00 AM' ORDER BY DateTimeofBilling