MYSQL 错误 1248 (42000)：每个派生表必须有自己的别名

Question

我一生都无法弄清楚为什么这是错误的

SELECT * FROM 
SELECT 
c.city_id,
p.prop_ynow_id,
p.propertyid,
p.prop_add_value,
p.name,
picture,
ifnull(p.address,'') as`location`,
ifnull(city,'')as`city`,
ifnull(ShortCut,'') as `state`,
ifnull(p.zip,'') as `zip`,
min(if(pr.minrent = 0,99999999,pr.minrent)) as minrent, 
max(pr.maxrent) as maxrent,
'' as service,
hood_id,
ifnull(p.phone,'') as `phone`,
latitude,
longitude,
min(CAST(pu.fullBath AS UNSIGNED)) as`minbath`,
max(CAST(pu.fullBath AS UNSIGNED)) as`maxbath`,
min(CAST(pu.Bed AS UNSIGNED)) as`minbed` ,
max(CAST(pu.Bed AS UNSIGNED)) as`maxbed`,
'' as url,
'' as source_id,
'' as source_name,
'' as addresscode,
'' as citycode,
'' as ctime,
'' as paid,
'' as similar_url,
'' as created_at,
'' as updated_at,
'' as city_name,
'' as service_listing_id

FROM 
wiki_city_list c join propertyinfo p on c.city_id=p.city 
join ynow_rentwiki.Property_Unitlayout pu on p.prop_ynow_id=pu.P_Ident
join (SELECT CAST(substring_index(if(Rent >0 ,Rent,RentLow),'.',1) AS UNSIGNED) as minrent, CAST(substring_index(if(Rent >0,Rent,Renthigh),'.',1) AS UNSIGNED) as maxrent,PRE_Ident,P_Ident,UNL_Ident,RTY_Ident from ynow_rentwiki.Property_rents where P_Ident in (3958, 4576, 4577) and (Rent!='' or (Rentlow!='' and Renthigh!='')) )  as pr on pu.UNL_Ident=pr.UNL_Ident
join state s on (p.state = s.stateid OR p.state = s.ShortCut ) 
WHERE 
pu.Status='Active'
and p.delete_date='0000-00-00'

GROUP BY 
c.city_id, p.prop_ynow_id

UNION 
SELECT 
c.city_id,
p.prop_ynow_id,
p.propertyid,
p.prop_add_value,
p.name,
picture,
ifnull(p.address,'') as`location`,
ifnull(city,'')as`city`,
ifnull(ShortCut,'') as `state`,
ifnull(p.zip,'') as `zip`,
min(if(pr.minrent = 0,99999999,pr.minrent)) as minrent, 
max(pr.maxrent) as maxrent,
'' as service,
hood_id,
ifnull(p.phone,'') as `phone`,
latitude,
longitude,
min(CAST(pu.fullBath AS UNSIGNED)) as`minbath`,
max(CAST(pu.fullBath AS UNSIGNED)) as`maxbath`,
min(CAST(pu.Bed AS UNSIGNED)) as`minbed` ,
max(CAST(pu.Bed AS UNSIGNED)) as`maxbed`,
'' as url,
'' as source_id,
'' as source_name,
'' as addresscode,
'' as citycode,
'' as ctime,
'' as paid,
'' as similar_url,
'' as created_at,
'' as updated_at,
'' as city_name,
'' as service_listing_id

FROM 
wiki_city_list c join propertyinfo p on c.city_id=p.city 
join ynow_rentwiki.Property_Unitlayout pu on p.prop_ynow_id=pu.P_Ident
join (SELECT CAST(substring_index(if(Rent >0 ,Rent,RentLow),'.',1) AS UNSIGNED) as minrent, CAST(substring_index(if(Rent >0,Rent,Renthigh),'.',1) AS UNSIGNED) as maxrent,PRE_Ident,P_Ident,UNL_Ident,RTY_Ident from ynow_rentwiki.Property_rents where P_Ident in (9744) and (Rent!='' or (Rentlow!='' and Renthigh!='')) )  as pr on pu.UNL_Ident=pr.UNL_Ident
join state s on (p.state = s.stateid OR p.state = s.ShortCut ) 
WHERE 
pu.Status='Active'
and p.delete_date='0000-00-00'

GROUP BY 
c.city_id, p.prop_ynow_id

UNION 
SELECT  
'' as prop_ynow_id, 
id as propertyid, 
0 as prop_add_value,
t.name as name,
'' as picture,  
t.address as location,
t.city as city, 
s.ShortCut as state, 
t.zip as zip,   
CAST(REPLACE(REPLACE(t.price,'$',''),',','') as UNSIGNED) as minrent, 
'' as maxrent,
t.service as service, 
'' as hood_id, 
'' as phone, 
t.latitude as latitude, 
t.longitude as longitude, 
t.bathrooms as minbath, 
'' as maxbath, 
t.bedrooms as minbed,
'' as maxbed,   
t.url as url,   
t.source_id as source_id, 
t.source_name as source_name, 
t.addresscode as addresscode, 
t.citycode as citycode, 
t.ctime as ctime, 
t.paid as paid,
t.similar_url as similar_url, 
t.created_at as created_at, 
t.updated_at as updated_at, 
SUBSTRING_INDEX(c.city_name,'_',1) as city_name,    
t.service_listing_id as service_listing_id

FROM LBCPrimary.third_party_properties as t, LBCPrimary.wiki_city_list as c, LBCPrimary.state as s
WHERE 
t.city in ( '230' ) 
and 
address <> '' and 
t.city = c.city_id and 
c.city_state = s.stateid

order by t.ctime 
desc
limit 46 as a limit 0,50

Answer 1

正如它所说的那样 - 每个派生表必须有一个别名。 SELECT a.* FROM (SELECT ....)a

更新。
这应该适合你：

SELECT xxx.* FROM 
(
    SELECT ....
    FROM ....
    UNION
    (
       SELECT ....
       FROM .....
       LIMIT 46
    )
    LIMIT 50
)xxx

Answer 2

查询的第一行

SELECT * FROM 

似乎是不必要的（因为所有三个 UNIONed 查询已经包含 SELECT 和 FROM 子句）。

删除这条不必要的线应该可以解决问题；或者，在第一行之后添加 ( ，并在末尾添加 ) QRYALIAS 也可以解决该问题。

Answer 3

选择 h.黑客_id，
h.姓名，
COUNT (c.challenge_id) AS 总数
来自黑客h，
挑战c
其中 h.hacker_id,
h.姓名
HAVING COUNT (c.challenge_id) IN (SELECT MAX (total))
FROM (SELECT COUNT () AS Total FROM Challenge GROUP BY hacker_id))
或 COUNT (c.challenge_id) IN
（选择总计
从
(SELECT COUNT () AS 总计
来自挑战
按 hacker_id 分组）
按总计分组
有计数（总计）=1)
按计数排序 (c.challenge_id) DESC, h.黑客_id；