Question

我正在尝试计算给定时间段内的登录次数，目前我的SQL查询仅显示至少登录一次的结果，我希望它甚至显示登录次数为零的结果。

我正在使用的查询：

SELECT c.FullName, COUNT(l.Id) 
FROM LoginsTable l JOIN UsersTable u ON u.Email = l.Email JOIN Organisations c ON c.Id = u.OrganisationId 
WHERE l.AttemptTime > "2019-10-01" AND l.AttemptTime < "2019-11-01" AND l.Success = 1 
GROUP BY c.Name
ORDER BY c.Name ASC;

Answer 1

您遇到了一些问题。首先，您需要使用RIGHT JOIN中的LoginsTable或对JOIN进行重新排序以将JOIN放在LoginsTable的最后，然后使用{{1} }。考虑到查询的性质，后者可能更有意义。其次，您需要在已LEFT JOIN设置为联接条件的表中的字段上放置任何条件，否则MySQL会将LEFT JOIN转换为LEFT JOIN（请参阅{{3} }。最后，您应该INNER JOIN GROUP BY中的相同字段。这应该起作用：

SELECT

Answer 2

我在这里发现2个问题：

您的分组依据未列在您的列中

日期条件使用双引号。

尝试以下查询。

SELECT c.FullName, COUNT(l.Id) 
FROM LoginsTable l 
LEFT JOIN UsersTable u ON u.Email = l.Email 
LEFT JOIN Organisations c ON c.Id = u.OrganisationId 
WHERE l.AttemptTime between '2019-10-01' AND '2019-11-01' AND l.Success = 1 
GROUP BY c.FullName
ORDER BY c.FullName ASC;

Answer 3

正如罗曼·霍克（Roman Hocke）所说，您需要按如下方式使用左连接：

SELECT c.FullName, COUNT(l.Id) 
FROM UsersTable u
JOIN Organisations c ON c.Id = u.OrganisationId 
LEFT JOIN LoginsTable l ON u.Email = l.Email 
WHERE l.AttemptTime > "2019-10-01" AND l.AttemptTime < "2019-11-01" AND l.Success = 1 
GROUP BY c.Name
ORDER BY c.Name ASC;

此外，您应该使用以下字段来修复组或选择该字段：SELECT c.Name或GROUP BY c.FullName ORDER BY c.FullName

编辑：尼克的答案就是一个。正如他所说的很好，您需要将条件放在左联接的on子句中。

SELECT c.FullName, COUNT(l.Id) 
FROM UsersTable u
JOIN Organisations c ON c.Id = u.OrganisationId 
LEFT JOIN LoginsTable l ON (u.Email = l.Email AND l.AttemptTime > "2019-10-01" AND l.AttemptTime < "2019-11-01" AND l.Success = 1)
GROUP BY c.FullName
ORDER BY c.FullName ASC;

显示结果也没有计数/零

3 个答案: