Question

我有2个表，我希望case语句子查询将返回值字符串值“ Rural”（如果不匹配）设置为“ Urban”（如果为true）。我需要查询表2来查看城市名称是否存在。

当前，我的查询始终返回错误值'RURAL'

, CASE --This CASE is identifying Urban communities as listed in the contract, and leaving the rest as Rural.

WHEN t1.city in (select top 1 city from t1
                 left join t2 on t2.city2 = t1.city
                where t2.city2 = t1.city 
                              ) 
            then 'Urban'
            ELSE 'Rural' 
          end as URBAN_RURAL

select    t1.city
    , t2.city2
    , CASE --This CASE is identifying Urban communities and leaving the rest as Rural.
WHEN Exists (select top 1 from t1 where t1.city = t2.city2) 
            then 'Urban'
            ELSE 'Rural' 
      end as Urban_Status; 

from t1  with (nolock)
    left join t2 with (nolock)
    on t1.city = t2.city

我想使用查找表，因为我们可能正在修改符合城市资格的内容，而只是从该表t2中添加或删除名称。因此，如果case语句可以返回Urban字符串值以匹配两个表中的结果，而Rural则返回不匹配项，那么将是理想的。

示例结果为

City    City2     UrbanRural
Orono   NULL      Rural
Bangor  Bangor    Urban
Machias Null      Rural

Answer 1

我想这就是你想要的。但是如果没有样本预期结果，就不能确定。

SELECT COALESCE(
         ( SELECT 'URBAN'
             FROM T2
            WHERE T1.city = T2.city
         ), 'RURAL' ) AS CITY_TYPE,
       T1.*
  FROM T1

Answer 2

如果您已经在t1子句中加入了t2和FROM，则根本不需要内联子查询。

SELECT 
  CASE 
    WHEN t2.city(2?) IS NULL THEN 'URBAN'
    ELSE 'RURAL'
  END AS URBAN_RURAL
FROM t1
LEFT JOIN t2 
  ON t1.city = t2.city

注意：NOLOCK已删除-如果您确实需要它，请放回去，但请参阅Bad habits : Putting NOLOCK everywhere

SQL select子查询基于case语句返回布尔值

2 个答案: