我希望能够为每个菜单返回 5 个菜单项.我试过这几个脚本,但没有运气.这是桌子
I want to be able to return 5 menuitem per menu. I've tried this several script but had no luck. here are the tables
menus
-------
menuid int()
profileName varchar(35)
menuitems
-----------
itemid int()
name varchar(40)
这是我现在拥有的.我收到以下脚本的错误消息.错误:子查询返回多于 1 行.
Here what I have now. I'm getting the error message with the script below. Error: Subquery returns more then 1 row.
SELECT m.profilename, name
FROM menus m
WHERE (SELECT name
from menuitems s
where m.menuid = s.menuid
limit 5)
非常感谢任何建议.
你必须为此使用副作用变量
You have to use side effecting variables for this
SELECT profilename, name
FROM
(
SELECT m.profilename, s.name,
@r:=case when @g=m.profilename then @r+1 else 1 end r,
@g:=m.profilename
FROM (select @g:=null,@r:=0) n
cross join menus m
left join menuitems s on m.menuid = s.menuid
) X
WHERE r <= 5
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