從串聯

Question

刪除charecters我與佈局如下表：

CREATE TABLE dbo.tbl (
    Ten_Ref VARCHAR(20) NOT NULL, 
    Benefit VARCHAR(20) NOT NULL 
); 

INSERT INTO dbo.tbl (Ten_Ref, Benefit) 
VALUES ('1', 'HB'), 
     ('1', 'WTC'), 
     ('1', 'CB'), 
     ('2', 'CB'), 
     ('2', 'HB'), 
     ('3', 'WTC'); 

我然後運行該代碼來執行轉換和連接（我需要的所有服務信息的一個領域」

with [pivot] as 

(
SELECT Ten_Ref 
,[HB] = (Select Benefit FROM tbl WHERE t.Ten_Ref = Ten_Ref and Benefit = 'HB') 
,[CB] = (Select Benefit FROM tbl WHERE t.Ten_Ref = Ten_Ref and Benefit = 'CB') 
,[WTC] = (Select Benefit FROM tbl WHERE t.Ten_Ref = Ten_Ref and Benefit = 'WTC') 
/*Plus 7 more of these*/ 

FROM tbl as t 

GROUP BY Ten_Ref 
) 

select p.ten_Ref 
     /*A concatenation to put them all in one field, only problem is you end up with loads of spare commas*/ 
     ,[String] = isnull (p.HB,0) + ',' + isnull (p.cb,'') + ',' + isnull (p.wtc,'') 

from [pivot] as p 

我的問題是不是每個ten_ref有所有優點的連接。

使用此代碼，其中還有一定的差距Ø r NULL然後我最終得到大量的雙逗號，例如'HB ,, WTC'

我怎樣才能得到它，所以它只有一個逗號，而不管每個租約有多少好處？

謝謝 - （第一篇文章）

Answer 1

你在找這樣的嗎？

SELECT A.Ten_Ref, 
     STUFF(CA.list,1,1,'') list 
FROM tbl A 
CROSS APPLY(
       SELECT ',' + Benefit 
       FROM tbl B 
       WHERE A.Ten_Ref = B.Ten_Ref 
       ORDER BY Benefit 
       FOR XML PATH('') 
      ) CA(list) 
GROUP BY A.ten_ref,CA.list 

結果：

Ten_Ref    list 
-------------------- ------------------ 
1     CB,HB,WTC 
2     CB,HB 
3     WTC 

或者，如果你真的想用支點和手工拼接，你可以這樣做：

SELECT Ten_Ref, 
     --pvt.*, 
     ISNULL(HB + ',','') + ISNULL(CB + ',','') + ISNULL(WTC + ',','') AS list 
FROM tbl 
PIVOT 
(
    MAX(Benefit) FOR Benefit IN([HB],[CB],[WTC]) 
) pvt