Question

我有以下查询，如果故事表中的每个mainMenu，subMenu和subMenu2列都有值，则会产生正确的结果。

如果其中任何一个为null，则不返回任何结果。我不想使用左连接，因为当连接菜单处于非活动状态时，它将返回结果。

SELECT * 
FROM stories s
JOIN menu m
    ON s.mainMenu = m.id 
    AND m.menu_active = 1 
JOIN submenu m2
    ON s.subMenu = m2.id
    AND m2.submenu_active = 1
JOIN submenu_L2 m3
    ON s.subMenu2 = m3.id
    AND m3.submenu_L2_active = 1
WHERE s.story_id = 198 
AND s.story_active = 1

如果故事表中的相关列不为空，是否有办法只加入表？

类似于：

SELECT * 
FROM stories s
if s.mainMenu IS NOT NULL then
JOIN menu m
    ON s.mainMenu = m.id 
    AND m.menu_active = 1 
end if
JOIN sub me...

Answer 1

我同意Mukus。听起来你想要LEFT JOIN。

LEFT JOIN将从表格“故事”中返回所有内容，并从表格“菜单”中返回他们加入的数据。

SELECT * 
FROM stories s
LEFT JOIN menu m
    ON s.mainMenu = m.id 
    AND m.menu_active = 1
LEFT JOIN sub me...

Answer 2

我认为left outer join子句中需要where更多逻辑：

SELECT * 
FROM stories s LEFT JOIN
     menu m
     ON s.mainMenu = m.id AND m.menu_active = 1 JOIN
     submenu m2
     ON s.subMenu = m2.id AND m2.submenu_active = 1 LEFT JOIN
     submenu_L2 m3
     ON s.subMenu2 = m3.id AND m3.submenu_L2_active = 1
WHERE s.story_id = 198 AND s.story_active = 1 AND
      (s.mainMenue is null or m.id is not null);

MySQL - 如果x不为null，则连接表

2 个答案: