这是我的SQL查询:
Select 'P'+(cast(p.Id as varchar)) as SolrId,
(Select lp.LocaleValue where LanguageId=3 and lp.EntityId=p.Id) as fr_Name ,
(Select lp.LocaleValue where LanguageId=2) as hi_Name,
(Select lp.LocaleValue where LanguageId=4) as nl_Name,* from Product p
LEft Join LocalizedProperty lp on EntityId=p.Id
答案 0 :(得分:1)
我只会使用条件聚合:
Select 'P' + (cast(p.Id as varchar(255))) as SolrId,
max(case when lp.LanguageId = 3 then lp.LocaleValue end) as fr_name,
max(case when lp.LanguageId = 2 then lp.LocaleValue end) as hi_name,
max(case when lp.LanguageId = 4 then lp.LocaleValue end) as nl_name
from Product p left join
LocalizedProperty lp
on lp.EntityId = p.Id
group by p.id;
答案 1 :(得分:0)
自己加入lp并给它4个不同的别名
select lp1.LocalValue, lp2.Localvalue from lp lp1
join lp lp2 on lp1.pId = lp2.pId
where
lp1.Languageid = 3 and lp2.Languageid = 2
答案 2 :(得分:0)
这很好用
Select 'P' + (cast(p.Id as varchar(255))) as SolrId,(cast(p.Name as varchar(MAX))) as Name,
max(case when lp.LanguageId = 3 then lp.LocaleValue end) as fr_name,
max(case when lp.LanguageId = 2 then lp.LocaleValue end) as hi_name,
max(case when lp.LanguageId = 4 then lp.LocaleValue end) as nl_name,
max(case when lp.LanguageId = 4 then lp.LocaleValue end)
from Product p left join
LocalizedProperty lp
on lp.EntityId = p.Id
group by p.id,p.Name