如何仅从昨天选择sql数据库?这是我的代码:
SELECT *, COUNT(visitors.usr_id) as usr_count FROM user, visitors WHERE visitors.usr_id = $usr_id GROUP BY $usr_id ORDER BY usr_count LIMIT 1我的数据库名是timein以及它输入插入的日期和时间的方式。以下是代码:
Database column : timein
Database insert looks like : 2012-9-6 9:11:35基本上,我希望能够选择和计数只从昨天。如何从昨天的sql数据库中选择COUNT?
发布于 2013-12-14 19:25:31
试试这个,为昨天添加一个条件,比如timein <= NOW() - INTERVAL 1 DAY , timein >= NOW() - INTERVAL 2 DAY;
SELECT *, COUNT(visitors.usr_id) as usr_count
FROM user, visitors
WHERE visitors.usr_id = $usr_id
AND timein <= NOW() - INTERVAL 1 DAY
AND timein >= NOW() - INTERVAL 2 DAY
GROUP BY $usr_id
ORDER BY usr_count
LIMIT 1或者您可以使用BETWEEN
SELECT *, COUNT(visitors.usr_id) as usr_count
FROM user, visitors
WHERE visitors.usr_id = $usr_id
AND timein BETWEEN NOW() - INTERVAL 1 DAY AND NOW() - INTERVAL 2 DAY;
GROUP BY $usr_id
ORDER BY usr_count
LIMIT 1发布于 2013-12-14 19:27:51
您的查询没有意义。您正在user和visitors之间进行交叉连接,然后只对访问者进行过滤。我怀疑你想要这样的东西:
SELECT *, COUNT(visitors.usr_id) as usr_count
FROM user join
visitors
on visitors.usr_id = user.usr_id
WHERE user.usr_id = $usr_id
GROUP BY $usr_id
ORDER BY usr_count
LIMIT 1;要获取昨天数据的where子句是:
WHERE date(timein) = date(NOW() - INTERVAL 1 DAY)或者,如果您有一个timein索引
WHERE timein >= date(NOW() - INTERVAL 1 DAY) and timein < date(NOW())https://stackoverflow.com/questions/20587230
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