I would like to retrive a TODAY'S data from the database, but I don't know how to do it. I would actually want to get the data from NOT the past 24 hours, I just want today's data (so based on the actual server time).
我想从数据库中检索一个今天的数据,但我不知道该怎么做。我实际上想从过去24小时获取数据,我只想要今天的数据(所以基于实际的服务器时间)。
I would also like to get data which was yesterday. Can anyone help me how to do it?
我还想获得昨天的数据。任何人都可以帮我怎么做?
Sample code:
示例代码:
"SELECT id FROM folk WHERE time = ???"
Thank you in advance!
先谢谢你!
5 个解决方案
#1
13
I think you are looking for this:
我想你正在寻找这个:
"SELECT id FROM folk WHERE DATE(time) = CURDATE()"
time must be a field in you table that holds a reference to the row.
time必须是表中包含对行的引用的字段。
update
更新
To get yesterdays additions:
为了获得昨天的补充:
"SELECT id FROM folk WHERE DATE(time) = CURDATE() - 1"
update 2
更新2
To get all additions this month:
要在本月获得所有新增内容:
"SELECT id FROM folk
WHERE MONTH(time) = MONTH(NOW()) AND YEAR(time) = YEAR(NOW())"
reference: http://dev.mysql.com/doc/refman/5.5/en/date-and-time-functions.html
参考:http://dev.mysql.com/doc/refman/5.5/en/date-and-time-functions.html
#2
3
SELECT id FROM folk WHERE DATE(time) = DATE(NOW());
SELECT id FROM folk WHERE DATE(time) = DATE(DATE_SUB(NOW(), INTERVAL 1 DAY));
provided that 'time' has a proper date-time type
只要'时间'有适当的日期时间类型
#3
1
Try something like this>>
试试这样的东西>>
SELECT id from folk WHERE DAY( date ) = EXTRACT(DAY from (NOW() - inTERVAL
1 DAY ) )
Refer this link http://www.webmasterworld.com/forum112/278.htm
请参阅此链接http://www.webmasterworld.com/forum112/278.htm
#4
1
As you are using timestamp:
在使用时间戳时:
"SELECT id FROM folk WHERE time >= ".mktime(0, 0, 0)
That will select all data since beginning today.
这将从今天开始选择所有数据。
If you want to get all date not for today, you would do
如果你想得到今天的所有日期,你会这样做
"SELECT id FROM folk WHERE time < ".mktime(0, 0, 0)
To select data from yesterday, you would do:
要从昨天选择数据,您可以:
"SELECT id FROM folk WHERE time < ".mktime(0, 0, 0)." AND time >= ".mktime(0, 0, 0, date('m'), date('d')-1, date('Y'))
If you were to use DATETIME, just for reference, it would be something like:
如果您使用DATETIME,仅供参考,它将类似于:
"SELECT id FROM folk WHERE time >= '".date('Y-m-d').' 00:00:00."'"
#5
0
Get current date by using something like following query
通过使用以下查询来获取当前日期
SELECT *
FROM your_table_name
WHERE DAY('.$colum_name.') = DAY("2018-05-02")';
#1
13
I think you are looking for this:
我想你正在寻找这个:
"SELECT id FROM folk WHERE DATE(time) = CURDATE()"
time must be a field in you table that holds a reference to the row.
time必须是表中包含对行的引用的字段。
update
更新
To get yesterdays additions:
为了获得昨天的补充:
"SELECT id FROM folk WHERE DATE(time) = CURDATE() - 1"
update 2
更新2
To get all additions this month:
要在本月获得所有新增内容:
"SELECT id FROM folk
WHERE MONTH(time) = MONTH(NOW()) AND YEAR(time) = YEAR(NOW())"
reference: http://dev.mysql.com/doc/refman/5.5/en/date-and-time-functions.html
参考:http://dev.mysql.com/doc/refman/5.5/en/date-and-time-functions.html
#2
3
SELECT id FROM folk WHERE DATE(time) = DATE(NOW());
SELECT id FROM folk WHERE DATE(time) = DATE(DATE_SUB(NOW(), INTERVAL 1 DAY));
provided that 'time' has a proper date-time type
只要'时间'有适当的日期时间类型
#3
1
Try something like this>>
试试这样的东西>>
SELECT id from folk WHERE DAY( date ) = EXTRACT(DAY from (NOW() - inTERVAL
1 DAY ) )
Refer this link http://www.webmasterworld.com/forum112/278.htm
请参阅此链接http://www.webmasterworld.com/forum112/278.htm
#4
1
As you are using timestamp:
在使用时间戳时:
"SELECT id FROM folk WHERE time >= ".mktime(0, 0, 0)
That will select all data since beginning today.
这将从今天开始选择所有数据。
If you want to get all date not for today, you would do
如果你想得到今天的所有日期,你会这样做
"SELECT id FROM folk WHERE time < ".mktime(0, 0, 0)
To select data from yesterday, you would do:
要从昨天选择数据,您可以:
"SELECT id FROM folk WHERE time < ".mktime(0, 0, 0)." AND time >= ".mktime(0, 0, 0, date('m'), date('d')-1, date('Y'))
If you were to use DATETIME, just for reference, it would be something like:
如果您使用DATETIME,仅供参考,它将类似于:
"SELECT id FROM folk WHERE time >= '".date('Y-m-d').' 00:00:00."'"
#5
0
Get current date by using something like following query
通过使用以下查询来获取当前日期
SELECT *
FROM your_table_name
WHERE DAY('.$colum_name.') = DAY("2018-05-02")';