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mysql - 通过忽略日期中的时间部分来显示数据

转载 作者:行者123 更新时间:2023-11-29 13:08:39 29 4
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我正在使用JasperReports/iReport来制作我的报告。我已经编写了一个查询,让客户端键入日期,但有 1 个问题,日期是与时间部分一起出现的。

如何忽略客户端输入数据的时间部分?

date                            rate2/2/2014  12:56:21.0000 PM      ABC   4/2/2014  12:56:21.0000 PM      EFG5/2/2014  12:56:21.0000 PM      HIJ...

I have assign from customer to design with the parameter of year,month, and day.In other words, when the customer key in the year,month,and day,the data will show this way:

2/2/2014  12:56:21.0000 PM      ABC

The query i have wrote is show below:

select date,rate 
from mytable
where rownum=1
and tran_dt<=(select date
from mytable
where year(date)=$P{year}
and month(date)=$P{month}
and day(date)=$P{day}
)
order by
date

我正在使用 pl/sql 查询...

最佳答案

SQL 服务器:

使用这个

 select convert(varchar(10), '2011-02-25 21:17:33.933', 120)

像这样

 Select DATE_FORMAT(date,'%y-%m-%d'),rate from mytable
where rownum=1 and tran_dt<=(select tran_dt from bi01_trandetail where
year(date)=$P{year} and month(tran_dt)=$P{month} and day(tran_dt)=$P{day})
order by date

MYSQL

使用

 DATE_FORMAT(date,'%y-%m-%d')


Select DATE_FORMAT(date,'%y-%m-%d'),rate from mytable
where rownum=1 and tran_dt<=(select tran_dt from bi01_trandetail where
year(date)=$P{year} and month(tran_dt)=$P{month} and day(tran_dt)=$P{day})
order by date

或者

SELECT DATE('2003-12-31 01:02:03');
'2003-12-31'



Select date(date),rate from mytable
where rownum=1 and tran_dt<=(select tran_dt from bi01_trandetail where
year(date)=$P{year} and month(tran_dt)=$P{month} and day(tran_dt)=$P{day})
order by date

您的新查询:

select date,rate 
from mytable
where rownum=1
and tran_dt<=(select DATE_FORMAT(date,'%y-%m-%d')
from mytable
where year(date)=$P{year}
and month(date)=$P{month}
and day(date)=$P{day}
)
order by
date

MYSQL DATE FUNCTION

关于mysql - 通过忽略日期中的时间部分来显示数据,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/22373465/

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