mysql - 如何使用 "LEFT ON"获取SQLite中的数据

Question

SELECT CaseFolder.CaseId, CaseFolder.CaseCoverImageFileName,
            CaseFolder.CaseName, count(Event.CaseId), CaseFolder.CaseColor,
            CaseFolder.IsMedicalCase, CaseFolder.StartDate, CaseFolder.EndDate
    FROM CaseFolder
    LEFT JOIN Event ON CaseFolder.CaseId = Event.CaseId
    WHERE Event.PersonId = 1 AND Event.IsDeleted = 0 AND CaseFolder.IsDeleted = 0
    GROUP BY CaseFolder.CaseId 

我按表Event 选择Count，并按表CaseFolder 选择另一个。主键 是 CaseId，但是我需要选择一个表中包含但另一个表中不包含的数据，此时必须添加“where”和“LEFT ON”。如果我删除sql代码WHERE Event.PersonId = 1 AND Event.IsDeleted = 0 AND CaseFolder.IsDeleted = 0 结果是正确的。因为 WHERE 包含不适合表CaseFolder 的条件。我该怎么办？

Answer 1

左连接中第二个表的条件应位于 on 子句中:

SELECT CaseFolder.CaseId, CaseFolder.CaseCoverImageFileName,
       CaseFolder.CaseName, count(Event.CaseId), CaseFolder.CaseColor,
       CaseFolder.IsMedicalCase, CaseFolder.StartDate, CaseFolder.EndDate
FROM CaseFolder LEFT JOIN
     Event
     ON CaseFolder.CaseId = Event.CaseId AND
        Event.PersonId = 1 AND Event.IsDeleted = 0 
WHERE CaseFolder.IsDeleted = 0;
GROUP BY CaseFolder.CaseId ;

否则，LEFT JOIN 会变成 INNER JOIN，因为条件失败并返回 NULL 值。请注意，使用表别名可以更轻松地编写和读取查询:

SELECT cf.CaseId, cf.CaseCoverImageFileName,
       cf.CaseName, count(e.CaseId), cf.CaseColor,
       cf.IsMedicalCase, cf.StartDate, cf.EndDate
FROM CaseFolder cf LEFT JOIN
     Event e
     ON cf.CaseId = e.CaseId AND
        e.PersonId = 1 AND e.IsDeleted = 0 
WHERE cf.IsDeleted = 0;
GROUP BY cf.CaseId ;