SQL SELECT ID WHERE具有相同ID的行具有不同的值

Question

我需要一些帮助来创建跨行的SQL语句。

SELECT SZ.Stammindex AS ID, sEbene1, sEbene2, sEbene3
FROM SuchbaumZuordnung SZ 
LEFT JOIN Suchbaum S SZ.gSuchbaumID = S.gID
WHERE (sEbene1 IN ('Test1') 
       AND (sEbene2 IN ('Test2') OR sEbene2 IS NULL) 
       AND sEbene3 IS NULL)

<< img src =“ https://image.soinside.com/eyJ1cmwiOiAiaHR0cHM6Ly9pLnN0YWNrLmltZ3VyLmNvbS94ZVlmZC5wbmcifQ==” alt =“这是我到目前为止得到的东西”>

正如您在屏幕快照中所看到的，我选择了ID = 10004和ID = 10005。但是实际上我只希望显示ID = 10005。如前所述，我正在尝试对行进行过滤。

我的目标是获取所有ID，其中所有条件都与“ AND”相关联，如下所示：

WHERE (sEbene1 IN ('Test1') 
       AND (sEbene2 IN ('Test2') *AND* sEbene2 IS NULL) 
       AND sEbene3 IS NULL)

但是不会返回任何内容。

希望你们能帮助我。

Answer 1

我怀疑您想要：

SELECT SZ.Stammindex AS ID
FROM SuchbaumZuordnung SZ 
WHERE EXISTS (SELECT 1
              FROM Suchbaum S
              WHERE SZ.gSuchbaumID = S.gID AND
                    S.sEbene1 IN ('Test1') AND
                    sEbene2 IN ('Test2')
             ) AND
       EXISTS (SELECT 1
              FROM Suchbaum S
              WHERE SZ.gSuchbaumID = S.gID AND
                    S.sEbene2 IS NULL AND
                    S.sEbene3 IS NULL
             );

这正在Suchbaum中寻找两个
different行，每个行都与条件之一匹配。

Answer 2

考虑到您只有3列要检查不同的行，似乎可以通过CTE和Windowed COUNT轻松实现：

WITH CTE AS(
    SELECT SZ.Stammindex AS ID,
           S.sEbene1, --Guessed the table alias
           S.sEbene2, --Guessed the table alias
           S.sEbene3, --Guessed the table alias
           COUNT(DISTINCT CONCAT(ISNULL(S.S.sEbene1,'-'),ISNULL(S.sEbene2,'-'),ISNULL(S.sEbene3,'-'))) OVER (PARTITION BY SZ.Stammindex) AS DistinctRows
    FROM SuchbaumZuordnung SZ 
         LEFT JOIN Suchbaum S ON SZ.gSuchbaumID = S.gID) --This was missing the ON in your sample
SELECT C.Stammindex,
       C.sEbene1,
       C.sEbene2,
       C.sEbene3
FROM CTE C
WHERE C.DistinctRows > 1;

如果纯粹是一个ID多于1行（可能是相同的，那么您可以只使用COUNT：
WITH CTE AS(
    SELECT SZ.Stammindex AS ID,
           S.sEbene1, --Guessed the table alias
           S.sEbene2, --Guessed the table alias
           S.sEbene3, --Guessed the table alias
           COUNT(*) OVER (PARTITION BY SZ.Stammindex) AS [Rows]
    FROM SuchbaumZuordnung SZ 
         LEFT JOIN Suchbaum S ON SZ.gSuchbaumID = S.gID)
SELECT C.Stammindex,
       C.sEbene1,
       C.sEbene2,
       C.sEbene3
FROM CTE C
WHERE C.[Rows] > 1;