是否可以在CASE中使用count，这是我的查询。给了我一个错误＃1111
SELECT t1.id,t1.name,t1.date_entered,t1.date_modified, t1.adresse,t1.adresse_state,t1.adresse_city,t1.adresse_postalcode,t1.hausnummer,t1.center_lat,t1.center_long, t3.name as street_name,t3.strasse,t3.primaerstr
FROM hi_haus as t1
LEFT JOIN hi_haus_hi_strasse as t2 ON (t2.hi_haus_id = t1.id)
LEFT JOIN hi_strasse as t3 ON (t3.id = t2.hi_strasse_id)
WHERE t1.adresse !='' AND t1.adresse IS NOT NULL AND t1.adresse NOT LIKE '%bau%' AND t1.id IN ('58e722d6-44f5-4e24-8861-5cd4a6f07cbc','4d0a79ee-decf-49b1-8b4b-a912b278ffec')
AND t3.primaerstr =
CASE
WHEN count(t1.id) > '1' THEN t3.primaerstr='0'
ELSE
t3.primaerstr = '1'
END
GROUP BY t1.id
如果id重复多次，我需要的是获得primaerstr'0'。
答案 0 :(得分：0)
试试这个
SELECT t1.id ,
t1.name ,
t1.date_entered ,
t1.date_modified ,
t1.adresse ,
t1.adresse_state ,
t1.adresse_city ,
t1.adresse_postalcode ,
t1.hausnummer ,
t1.center_lat ,
t1.center_long ,
t3.name AS street_name ,
t3.strasse ,
t3.primaerstr
FROM hi_haus AS t1 ,
( SELECT id ,
COUNT(*) AS count
FROM hi_haus
GROUP BY id
) AS subquery
LEFT JOIN hi_haus_hi_strasse AS t2 ON ( t2.hi_haus_id = t1.id )
LEFT JOIN hi_strasse AS t3 ON ( t3.id = t2.hi_strasse_id )
WHERE t1.adresse != ''
AND t1.adresse IS NOT NULL
AND t1.adresse NOT LIKE '%bau%'
AND t1.id IN ( '58e722d6-44f5-4e24-8861-5cd4a6f07cbc',
'4d0a79ee-decf-49b1-8b4b-a912b278ffec' )
AND (( t3.primaerstr = 0
AND t1.id = subquery.id
AND subquery.count > 1 )
OR (t3.primaerstr = 1
AND t1.id = subquery.id
AND subquery.count = 0)
)
GROUP BY t1.id
答案 1 :(得分：0)
SELECT t1.id，t1.name，t1.date_entered，t1.date_modified，t1.adresse，t1.adresse_state，t1.adresse_city，t1.adresse_postalcode，t1.hausnummer，t1.center_lat，t1.center_long，t3。 name as street_name，t3.id，t3.strasse，t3.primaerstr FROM hi_haus as t1 LEFT JOIN hi_haus_hi_strasse as t2 ON（t2.hi_haus_id = t1.id）LEFT JOIN hi_strasse as t3 ON（t3.id = t2.hi_strasse_id）WHERE t1.adresse！=＆＃39;＆＃39; AND t1.adresse IS NOT NULL且t1.adresse NOT LIKE＆＃39;％bau％＆＃39; AND t1.id IN（＆＃39; 58e722d6-44f5-4e24-8861-5cd4a6f07cbc＆＃39;，＆＃39; 5d492f56-c26e-4c03-b3ec-e1d9b0c6f060＆＃39;，＆＃39; fc87b46a-5fe8-4ed0- b3f4-f6fbb0d6919d＆＃39;）AND t3.primaerstr！=（CASE WHEN（SELECT COUNT（hi_haus_id）FROM hi_haus_hi_strasse WHERE hi_haus_id = t1.id）＆gt; 1那么1 ELSE＆＃39; END）ORDER BY t1 .id，t1.date_modified DESC LIMIT 30;