用“-”分隔列（用于 xml 路径）答案

【问题标题】：separating a column with '-' (for xml path)用“-”分隔列（用于 xml 路径）
【发布时间】：2020-11-20 17:57:08
【问题描述】：

我想用“-”分隔查询中的“KaliteKodu”列，但它不起作用。在查询结果中，UrunID = 19241 的质量代码将如下所示

ProsesID  UrunID  Miktar           Tarih            LotNo      KaliteKodu
   1       19241   216     2020-11-20 00:00:00     FHM010901    A-A1-A2

查询：

select t.ProsesID,t.UrunID,sum(t.Miktar_Ad) Miktar,t.Tarih,t.LotNo,


KaliteKodu = stuff (
(select '-' + t.KaliteKodu 
    from IT07_GerCikanMalzeme GCM2 
    INNER JOIN IT06_StokKarti SK ON GCM2.UrunID = SK.UrunID
    INNER JOIN IT06_LotM LM ON SK.LotID = LM.LotID
    inner join IT01_KaliteKodlari KK ON GCM2.KaliteID = KK.KaliteID
    where GCM2.ProsesID = 2 and convert(date,GCM2.Tarih) = convert(date,GETDATE()) AND 
    GCM2.KaliteID in (select KaliteID from IT01_KaliteKodlari where KaliteGrupID = 1 and ITActive = 1) and 
    (LM.LotNo not in  (select m.LotNo from ZKF_IT01_FinKonM m where ProsesID = 2 and m.LotNo is not null)) AND GCM2.UrunID = t.UrunID
    FOR XML PATH (''))
         , 1, 1, '')


from
(
select GCM.ProsesID,GCM.UrunID,GCM.Miktar_Ad,GCM.Tarih,LM.LotNo , KK.KaliteKodu



    from IT07_GerCikanMalzeme GCM 
    INNER JOIN IT06_StokKarti SK ON GCM.UrunID = SK.UrunID
    INNER JOIN IT06_LotM LM ON SK.LotID = LM.LotID
    inner join IT01_KaliteKodlari KK ON GCM.KaliteID = KK.KaliteID
    where GCM.ProsesID = 2 and convert(date,GCM.Tarih) = convert(date,GETDATE()) AND 
    GCM.KaliteID in (select KaliteID from IT01_KaliteKodlari where KaliteGrupID = 1 and ITActive = 1) and 
    (LM.LotNo not in  (select m.LotNo from ZKF_IT01_FinKonM m where ProsesID = 2 and m.LotNo is not null))
    --group by ProsesID,GCM.UrunID,lm.LotNo,GCM.Tarih, KK.KaliteKodu

    ) t

    group by t.ProsesID,t.UrunID,t.Tarih,t.LotNo,t.KaliteKodu
        group by t.ProsesID,t.UrunID,t.Tarih,t.LotNo,t.KaliteKodu

【问题讨论】：

请不要通过破坏您的帖子为他人增加工作量。通过在 Stack Exchange 网络上发帖，您已在 CC BY-SA 4.0 license 下授予 Stack Exchange 分发该内容的不可撤销的权利（即无论您未来的选择如何）。根据 Stack Exchange 政策，帖子的非破坏版本是分发的版本。因此，任何破坏行为都将被撤销。如果您想了解更多关于删除帖子的信息，请参阅：How does deleting work?

标签： sql sql-server for-xml-path

【解决方案1】：

有一个错过的加入像

FROM TE t2 WHERE t.ProsesID = t2.ProsesID

使用的脚本

    --CREATE TABLE TE (ProsesID int, UrunID int, Miktar int, tarih DateTime, LotNo varchar(20), KaliteKodu varchar(3))
    --INSERT INTO TE VALUES (2, 19241, 72, '2020-11-20', 'fhm010901', 'A')
    --INSERT INTO TE VALUES (2, 19241, 72, '2020-11-20', 'fhm010901', 'A1')
    --INSERT INTO TE VALUES (2, 19241, 72, '2020-11-20', 'fhm010901', 'A2')
    --INSERT INTO TE VALUES (2, 19241, 72, '2020-11-20', 'fhm010901', 'A')


SELECT t.ProsesID, t.UrunID, t.Tarih, t.LotNo, KaliteKodu = stuff((
                SELECT '-' + t2.KaliteKodu
                FROM (select distinct t3.KaliteKodu, t3.ProsesID from TE t3 WHERE t.ProsesID = t3.ProsesID ) t2
                FOR XML PATH('')
                ), 1, 1, '')
    FROM TE t
    GROUP BY t.ProsesID, t.UrunID, t.Tarih, t.LotNo

TE 等于你的代码

from
(
select GCM.ProsesID,GCM.UrunID,GCM.Miktar_Ad,GCM.Tarih,LM.LotNo , KK.KaliteKodu
from IT07_GerCikanMalzeme GCM 
INNER JOIN IT06_StokKarti SK ON GCM.UrunID = SK.UrunID
INNER JOIN IT06_LotM LM ON SK.LotID = LM.LotID
inner join IT01_KaliteKodlari KK ON GCM.KaliteID = KK.KaliteID
where GCM.ProsesID = 2 and convert(date,GCM.Tarih) = convert(date,GETDATE()) AND 
GCM.KaliteID in (select KaliteID from IT01_KaliteKodlari where KaliteGrupID = 1 and ITActive = 1) and 
(LM.LotNo not in  (select m.LotNo from ZKF_IT01_FinKonM m where ProsesID = 2 and m.LotNo is not null))
--group by ProsesID,GCM.UrunID,lm.LotNo,GCM.Tarih, KK.KaliteKodu

) t

结果：

+----------+--------+-------------------------+-----------+------------+
| ProsesID | UrunID | Tarih                   | LotNo     | KaliteKodu |
+----------+--------+-------------------------+-----------+------------+
| 2        | 19241  | 2020-11-20 00:00:00.000 | fhm010901 | A-A1-A2    |
+----------+--------+-------------------------+-----------+------------+

【讨论】：

@Emre Yıldırım：TE 在您的代码中等于 t，因为您的代码是从另一个超出问题主题的查询生成 t 的。您的问题只需在我在答案开头提到的 JOIN 中调用两次 t 即可解决
能完整分享一下代码吗？坦率地说，我有点困惑:(
@Emre Yıldırım 等待它，将尝试从您的查询中提取示例并编辑我的帖子
我按照你说的做了，成功了，但是出现了一个新问题。你可以看到我编辑的帖子:))
@Emre Yıldırım：请解释一下新问题，从您的编辑中无法理解。

【解决方案2】：

根据您的 SQL Server 版本，您可以尝试以下任一方法：

DECLARE @Data table ( KaliteKodu varchar(50), UrunID int );
INSERT INTO @Data VALUES
    ( 'A', 19241 ), ( 'A1', 19241 ), ( 'A2', 19241 );

-- Using STRING_AGG with SQL Server 2017+ -- 
SELECT
    UrunID,
    STRING_AGG( KaliteKodu, '-' ) AS KaliteKodu
FROM @Data
WHERE
    UrunID = 19241
GROUP BY
    UrunID;

-- Pre SQL Server 2017 --
SELECT DISTINCT
    d.UrunID,
    vals.KaliteKodu
FROM @Data AS d
OUTER APPLY (
    
    SELECT STUFF (
        ( SELECT '-' + KaliteKodu AS "text()" FROM @Data x WHERE x.UrunID = d.UrunID FOR XML PATH( '' ) ),
        1,
        1,
        ''
    ) AS KaliteKodu

) AS vals
WHERE
    UrunID = 19241;

两个例子都返回：

+--------+------------+
| UrunID | KaliteKodu |
+--------+------------+
|  19241 | A-A1-A2    |
+--------+------------+

【讨论】：