sql - 如果记录存在于 A 中则返回记录，否则如果记录存在于 B 中则返回记录，否则返回 null

Question

我试图用 SQL 表达这个想法

If record exist in A then return record else if record exist in B return record ELSE NULL

示例:选择队长(如果存在)，否则选择队中最年长的球员

WITH Captains_CTE (TeamName, PlayerName)
AS (
    SELECT TeamName, PlayerName FROM TeamPlayer WHERE IsCaptain = 1
), 
OldestPlayer_CTE (TeamName, PlayerName)
AS (
    SELECT TeamName, PlayerName FROM TeamPlayer AS tp
    INNER JOIN ( 
        SELECT TeamName, MAX(PlayerAge) AS MaxAge
        FROM TeamPlayer 
        GROUP BY TeamName
    ) AS old ON old.TeamName = tp.TeamName AND old.MaxAge = tp.PlayerAge

)
SELECT  CASE 
            WHEN Captains_CTE.TeamName IS NULL THEN OldestPlayer_CTE.TeamName
            ELSE Captains_CTE.TeamName 
        END AS TeamName,
        CASE 
            WHEN Captains_CTE.PlayerName IS NULL THEN OldestPlayer_CTE.PlayerName
            ELSE Captains_CTE.PlayerName 
        END AS PlayerName
FROM Captains_CTE
FULL OUTER JOIN OldestPlayer_CTE ON Captains_CTE.TeamName = OldestPlayer_CTE.TeamName

我可以通过对 A 和 B 进行外连接，然后为我希望返回的每一列构建一个 CASE 语句来做到这一点，但必须有更好的方法。 (是的，我的示例查询对年龄的唯一性做出了一些可怕的假设)

Answer 1

使用不存在:

select a.*
from a
union all
select b.*
from b
where not exists (select 1 from a where a.? = b.?);  -- "?" is for the column that specifies whether the record exists in A