如何只使用 Microsoft SQL Server 中的日期查询 dATeTIME 字段？

Try this

 select * from test where Convert(varchar, date,111)= '03/19/2014'

use range, or DateDiff function

 select * from test
where date between '03/19/2014' and '03/19/2014 23:59:59'

or

 select * from test
where datediff(day, date, '03/19/2014') = 0

Other options are:

1. If you have control over the database schema, and you don't need the time data, take it out.

2. or, if you must keep it, add a computed column attribute that has the time portion of the date value stripped off...

Alter table Test Add DateOnly As DateAdd(day, datediff(day, 0, date), 0)

or, in more recent versions of SQL Server...

Alter table Test Add DateOnly As Cast(DateAdd(day, datediff(day, 0, date), 0) as Date)

then, you can write your query as simply:

select * from test
where DateOnly = '03/19/2014'

You can use this approach which truncates the time part:

select * from test
where convert(datetime,'03/19/2014',102) = DATEADD(dd, DATEDIFF(dd, 0, date), 0)

you can try this

select * from test where DATEADD(dd, 0, DATEDIFF(dd, 0, date)) = '03/19/2014';

Simple answer;

select * from test where cast ([date] as date) = '03/19/2014';

I am using MySQL 5.6 and there is a DATE function to extract only the date part from date time. So the simple solution to the question is -

 select * from test where DATE(date) = '2014-03-19';

http://dev.mysql.com/doc/refman/5.6/en/date-and-time-functions.html

-- Reverse the date format
-- this false:
select * from test where date = '28/10/2015'
-- this true:
select * from test where date = '2015/10/28'

There is a problem with dates and languages and the way to avoid it is asking for dates with this format YYYYMMDD.

This way below should be the fastest according to the link below. I checked in SQL Server 2012 and I agree with the link.

select * from test where date >= '20141903' AND date < DATEADD(DAY, 1, '20141903');

• Bad habits to kick : mis-handling date / range queries

select * from test
where date between '03/19/2014' and '03/19/2014 23:59:59'

This is a realy bad answer. For two reasons.

1. What happens with times like 23.59.59.700 etc. There are times larger than 23:59:59 and the next day.

2. The behaviour depends on the datatype. The query behaves differently for datetime/date/datetime2 types.

Testing with 23:59:59.999 makes it even worse because depending on the datetype you get different roundings.

select convert (varchar(40),convert(date      , '2014-03-19 23:59:59.999'))
select convert (varchar(40),convert(datetime  , '2014-03-19 23:59:59.999'))
select convert (varchar(40),convert(datetime2 , '2014-03-19 23:59:59.999'))

-- For date the value is 'chopped'. -- For datetime the value is rounded up to the next date. (Nearest value). -- For datetime2 the value is precise.

Test this query.

SELECT *,DATE(chat_reg_date) AS is_date,TIME(chat_reg_time) AS is_time FROM chat WHERE chat_inbox_key='$chat_key'
ORDER BY is_date DESC, is_time DESC

Simply use this in your WHERE clause.

The "SubmitDate" portion below is the column name, so insert your own.

This will return only the "Year" portion of the results, omitting the mins etc.

Where datepart(year, SubmitDate) = '2017'

select * from invoice where TRANS_DATE_D>= to_date  ('20170831115959','YYYYMMDDHH24MISS')
and TRANS_DATE_D<= to_date  ('20171031115959','YYYYMMDDHH24MISS');

This works for me for MS SQL server:

select * from test
where
year(date) = 2015
and month(date) = 10
and day(date)= 28 ;

select *, cast ([col1] as date) <name of the column> from test where date = 'mm/dd/yyyy'

"col1" is name of the column with date and time
<name of the column> here you can change name as desired

select *
from invoice
where TRUNC(created_date) <=TRUNC(to_date('04-MAR-18 15:00:00','dd-mon-yy hh24:mi:ss'));

SELECT * FROM test where DATEPART(year,[TIMESTAMP]) = '2018'  and  DATEPART(day,[TIMESTAMP]) = '16' and  DATEPART(month,[TIMESTAMP]) = '11'

use this

select * from TableName where DateTimeField > date() and  DateTimeField < date() + 1

use trunc(column).

select * from test t where trunc(t.date) = TO_DATE('2018/06/08', 'YYYY/MM/DD')