如何选择记录从过去24小时使用SQL?

我正在寻找一个where子句，可用于检索过去24小时的记录?

当前回答

在MySQL中:

SELECT  *
FROM    mytable
WHERE   record_date >= NOW() - INTERVAL 1 DAY

在SQL Server中:

SELECT  *
FROM    mytable
WHERE   record_date >= DATEADD(day, -1, GETDATE())

在Oracle中:

SELECT  *
FROM    mytable
WHERE   record_date >= SYSDATE - 1

在PostgreSQL:

SELECT  *
FROM    mytable
WHERE   record_date >= NOW() - '1 day'::INTERVAL

红移:

SELECT  *
FROM    mytable
WHERE   record_date >= GETDATE() - '1 day'::INTERVAL

SQLite:

SELECT  *
FROM    mytable
WHERE   record_date >= datetime('now','-1 day')

在MS Access中:

SELECT  *
FROM    mytable
WHERE   record_date >= (Now - 1)

其他回答

如果考虑的时间戳是UNIX时间戳您需要首先将UNIX时间戳(例如1462567865)转换为mysql时间戳或数据

SELECT * FROM `orders` WHERE FROM_UNIXTIME(order_ts) > DATE_SUB(CURDATE(), INTERVAL 1 DAY)

SELECT * 
FROM tableName 
WHERE datecolumn >= dateadd(hour,-24,getdate())

在postgres中，假设你的字段类型是时间戳:

Select * from date_field > (now() - interval '24小时');

在MySQL中:

SELECT  *
FROM    mytable
WHERE   record_date >= NOW() - INTERVAL 1 DAY

在SQL Server中:

SELECT  *
FROM    mytable
WHERE   record_date >= DATEADD(day, -1, GETDATE())

在Oracle中:

SELECT  *
FROM    mytable
WHERE   record_date >= SYSDATE - 1

在PostgreSQL:

SELECT  *
FROM    mytable
WHERE   record_date >= NOW() - '1 day'::INTERVAL

红移:

SELECT  *
FROM    mytable
WHERE   record_date >= GETDATE() - '1 day'::INTERVAL

SQLite:

SELECT  *
FROM    mytable
WHERE   record_date >= datetime('now','-1 day')

在MS Access中:

SELECT  *
FROM    mytable
WHERE   record_date >= (Now - 1)

MySQL:

SELECT * 
FROM table_name
WHERE table_name.the_date > DATE_SUB(NOW(), INTERVAL 24 HOUR)

时间间隔可以是年、月、日、小时、分、秒

比如，在最后十分钟

SELECT * 
FROM table_name
WHERE table_name.the_date > DATE_SUB(NOW(), INTERVAL 10 MINUTE)

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