基于 case 语句返回布尔值的 SQL 选择子查询

Question

我有 2 个表，如果不匹配，我希望 case 语句子查询设置返回值字符串值“Rural”，如果为真，则设置“Urban”。我需要查询来查找表 2 以查看城市名称是否存在。

目前我的查询总是返回错误值“农村”

, 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条件的城市，并且只是在该表中添加或删除名称 t2。因此，如果 case 语句可以为两个表中的匹配结果返回 Urban 字符串值，并且为不匹配的结果返回农村字符串值，那将是理想的。

样本结果是

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

Answer 1

我想这就是你想要的。但如果没有样本预期结果，则不确定。

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 FROM 子句中加入 t1 和 t2，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 - 如果您真的需要它，请将其放回原处，但请参阅 Bad habits : Putting NOLOCK everywhere