MYSQL ERROR 1248 (42000): Every derived table must have its own alias

Question

I cant for the life of me figure out why this is erroring

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

Your query start with SELECT * FROM SELECT - may you forget brackets? and should be SELECT * FROM (SELECT .... — Michael Pakhantsov, Jul 29 '10 at 15:06

a1ex07 · Accepted Answer · 2010-07-29T15:29:28.553

42

It means exactly what it says - each derived table must have an alias. SELECT a.* FROM (SELECT ....)a

Update. This should work for you:

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

edited Jul 29 '10 at 15:29

answered Jul 29 '10 at 15:05

a1ex07

36,826
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1

I am not sure what you mean...i tried to add it there are but now i get ERROR 1064 (42000): You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'as a limit 0,50)' at line 144 – Matt Elhotiby Jul 29 '10 at 15:09
Sorry, I didn't notice that one of your table has already had alias a. Use a different alias - `SELECT xxxx.* FROM (SELECT ....)xxxx` – a1ex07 Jul 29 '10 at 15:13
And when you use alias for `SELECT ... FROM t1 JOIN t2 JOIN t3...` you should also enclose SELECT into brackets (`(SELECT ... FROM t1 JOIN t2 JOIN t3 ...) [AS] a`). It why it complaints on near 'as a limit 0,50)' at line 144" – a1ex07 Jul 29 '10 at 15:19

score 4 · Answer 2 · answered Jul 29 '10 at 15:42

The very first line of your query is

SELECT * FROM

which appears to be unnecessary (as all three UNIONed queries already include SELECT and FROM clauses).

Removing this unnecessary line should resolve the problem; alternatively, adding a ( just after the first line, and adding a ) QRYALIAS at the end would also resolve the problem.

score -1 · Answer 3 · answered Feb 07 '22 at 12:30

-1

SELECT h. hacker_id, h. name, COUNT (c. challenge_id) AS total FROM hackers h, challenges c WHERE h.hacker_id, h. name HAVING COUNT (c. challenge_id) IN (SELECT MAX (total) FROM (SELECT COUNT () AS total FROM challenge GROUP BY hacker_id)) OR COUNT (c. challenge_id) IN (SELECT total FROM (SELECT COUNT () AS total FROM challenges GROUP BY hacker_id) GROUP BY total HAVING COUNT (total)=1) ORDER BY COUNT (c. challenge_id) DESC, h. hacker_id;

answered Feb 07 '22 at 12:30

Mishael Timothy

1

1

As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers [in the help center](/help/how-to-answer). – Community Feb 07 '22 at 15:03

MYSQL ERROR 1248 (42000): Every derived table must have its own alias

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