Question

我有一个LIKES的MySQL表（比如ID，userID，objectID，likeDate），我希望能够计算在有问题的用户之后所做的所有'喜欢'。

通常我会得到日期：

SELECT likeDate FROM LIKES WHERE userID = <logged in user's ID>

然后查找所有日期并计算返回的行（或使用mysql COUNT），如下所示：

SELECT * FROM LIKES WHERE likeDate > <given date>

但是，我确信有一种方法可以在一个查询中执行此操作，而不是对数据库进行两次调用。有人可以帮忙吗？

由于

Answer 1

将第一个查询的结果直接输入第二个查询：

SELECT COUNT(*)
FROM LIKES
WHERE likeDate > (
    SELECT max(likeDate)
    FROM LIKES
    WHERE userID = <logged in user's ID>
)

但请注意，您需要在第一次查询中添加max()。

此查询应该是获得答案的最快方式。为确保最佳性能，请在userID和likeDate上添加索引：

create index likes_userId on likes(userID);
create index likes_likeDate on likes(likeDate);

Answer 2

SELECT l1.likeDate, 
    (SELECT COUNT(1) FROM LIKES l2 WHERE l2.likeDate > l1.likeDate) AS likesAfter
FROM LIKES l1
WHERE userID = ?
GROUP BY l1.likeDate

或作为加入，

SELECT l1.likeDate, COUNT(1)
FROM LIKES l1
LEFT OUTER JOIN LIKES l2 ON l2.likeDate > l1.likeDate
WHERE userID = ?
GROUP BY l1.likeDate

Answer 3

SELECT * FROM LIKES WHERE likeDate > 
IFNULL((SELECT max(likeDate) FROM LIKES WHERE userID = <logged in user's ID> 
                                  adn  objectId=<question's Id>),0)
and objectId=<question's Id>

提高我的SQL效率

3 个答案: