Question

有人可能知道如何从SQL中获取一个逗号分隔的列表，而不是重复 - 这有点难以解释。让我举个例子。

我有一张发票清单+它所属的货物，如下表所示：

InvoiceNumber ShipmentNumber
0180376000    1stShipment
0180376005    1stShipment
0180376003    1stShipment
0180375997    1stShipment
0180375993    1stShipment

此列表需要分为主InvoiceNumbers，后跟剩余发票编号的右2位数。结果应类似于以下内容。

01803760, 00, 05, 03, 01803759, 97, 93

此时我可以很容易地得到逗号分隔列表，但无法弄清楚如何在它所属的每个相应发票之后定位2位数。

任何关于如何做到这一点的建议都会很棒!!

Answer 1

尝试这样

DECLARE @tbl TABLE(InvoiceNumber VARCHAR(100));
INSERT INTO @tbl VALUES
 ('0180376000')
,('0180376005')
,('0180376003')
,('0180375997')
,('0180375993');

WITH CutInTwo AS
(
    SELECT LEFT(InvoiceNumber,8) AS Number
          ,RIGHT(InvoiceNumber,2) AS SubNumber
    FROM @tbl
)
,OnlyMainNumbers AS
(
    SELECT DISTINCT Number
    FROM CutInTwo
)
SELECT y.Number + ' ' + (STUFF((SELECT ', ' + x.SubNumber FROM CutInTwo AS x WHERE x.Number=y.Number FOR XML PATH('')),1,2,''))
FROM OnlyMainNumbers AS y

如果你真的需要一个单独的字符串，你可以像这样包装最终的选择：

SELECT STUFF(
(
    SELECT ', ' + y.Number + ', ' + (STUFF((SELECT ', ' + x.SubNumber FROM CutInTwo AS x WHERE x.Number=y.Number FOR XML PATH('')),1,2,''))
    FROM OnlyMainNumbers AS y
    FOR XML PATH('')
),1,2,'')

Answer 2

试试这个

DECLARE @tbl TABLE (InvoceNumber NVARCHAR(50))

INSERT INTO @tbl VALUES  ('0180376000')
INSERT INTO @tbl VALUES  ('0180376005')
INSERT INTO @tbl VALUES  ('0180376003')
INSERT INTO @tbl VALUES  ('0180375997')
INSERT INTO @tbl VALUES  ('0180375993')

SELECT  
    (
        SELECT
             A.InvoceNumber + ', ' 
        FROM
        (
            SELECT DISTINCT LEFT(InvoceNumber, LEN(InvoceNumber) - 2) InvoceNumber  FROM @tbl
            UNION ALL
            SELECT RIGHT(InvoceNumber, 2) InvoceNumber FROM @tbl
        ) A
        FOR XML PATH ('')
    ) Invoce

输出：

01803759, 01803760, 00, 05, 03, 97, 93,

如果订单很重要。

SELECT
(
    SELECT
        Result.InvoceNumber + ', ' + Result.Invo
    FROM
    (
        SELECT
            A.InvoceNumber,
            (
                SELECT
                     IA.Invo   + ', ' AS [text()]
                FROM
                (
                    SELECT DISTINCT LEFT(InvoceNumber, LEN(InvoceNumber) - 2) InvoceNumber, RIGHT(InvoceNumber, 2) AS Invo  FROM @tbl
                ) IA
                WHERE
                    IA.InvoceNumber = A.InvoceNumber
                FOR XML PATH ('')
            ) Invo
        FROM
        (
            SELECT DISTINCT LEFT(InvoceNumber, LEN(InvoceNumber) - 2) AS InvoceNumber FROM @tbl
        ) A
        GROUP BY 
            A.InvoceNumber
    ) Result
    FOR XML PATH ('')
) S

输出：

01803759, 93, 97, 01803760, 00, 03, 05,

完整发票号+逗号分隔SQL列表（TSQL）

2 个答案: