我想做的事情:
选择date = 2012-10-14的行,然后显示该行之后的4行。所以从这个列表中
2012-10-12 column #2
2012-10-13 column #2
2012-10-14 was very sunny.
2012-10-15 rained all day.
2012-10-16 whatever.
2012-10-17 column #2
2012-10-18 column #2
2012-10-19 rained all day.
2012-10-20 whatever.
2012-10-21 column #2
2012-10-22 column #2
它会返回:
2012-10-14 was very sunny.
2012-10-15 rained all day.
2012-10-16 whatever.
2012-10-17 column #2
2012-10-18 column #2
感谢您的帮助。
PS:在数据库中,周末没有数据,所以有些日期会丢失。
答案 0 :(得分:5)
SELECT * FROM my_table WHERE date >= '2012-10-14' ORDER BY date LIMIT 5
在sqlfiddle上查看。
答案 1 :(得分:1)
此版本适用于日期范围,而不是固定数量的行。目前还不完全清楚你想要哪一个。
select * from mytable where date_diff(date,'2012-10-14') <= 4 and date >= '2012-10-14';