Sql server 合并/合并多列在一行中的唯一值

尝试在SSMS中使用单个查询组合多个列和行。下面是我正在处理的更复杂表的一个简单版本：

以下是我当前的查询：

CREATE TABLE Table1
([C1] varchar(50), [C2] varchar(50), [C3] varchar(50))
;

INSERT INTO Table1
([C1], [C2], [C3])
VALUES
('F92', 'Game1', 'b100'),
('F92', 'Game1', 'b200'),
('F92', 'Game2', 'C200'),
('F92', 'Game2', 'D400')
;



SELECT
 C1,C2,
 STUFF(
     (SELECT ', ' + C3
      FROM Table1
      WHERE C1 = a.C1 AND C2 = a.C2
      FOR XML PATH (''))
      , 1, 1, '')  AS NamesList
FROM Table1 AS a
GROUP BY C1,C2

drop table table1

我打算得到如下结果：

C1   | Namelist
F92  | Game1 b100, b200 Game2 c200, d400

---

这在单个查询中是否可行？

您确定可以做到这一点。您可以使用cte获取已获取的分隔列表。然后将结果转换为另一个分隔列表。但您可能需要一些东西作为组之间的分隔符，而不是空格

with FirstPass as
(
    SELECT
     C1,
     C2 + ' ' + STUFF(
         (SELECT ', ' + C3
          FROM Table1
          WHERE C1 = a.C1 AND C2 = a.C2
          FOR XML PATH (''))
          , 1, 1, '')  AS NamesList
    FROM Table1 AS a
    GROUP BY C1,C2
)

select C1
    , NameList = stuff((select ' ' + fp2.NamesList
        from FirstPass fp2
        where fp.C1 = fp2.C1
        FOR XML PATH('')), 1, 1, '')
from FirstPass fp
group by fp.C1

返回：

F92 |游戏1 b100、b200游戏2 C200、D400

--编辑--

对于需要按计数对C2进行排序的新要求，您只需添加一点聚合即可。这与原来的没有多大变化

with FirstPass as
(
    SELECT
     C1,
     C2 + ' ' + STUFF(
         (SELECT ', ' + C3
          FROM Table1
          WHERE C1 = a.C1 AND C2 = a.C2
          FOR XML PATH (''))
          , 1, 1, '')  AS NamesList
    , count(*) as GameCount
    FROM Table1 AS a
    GROUP BY C1,C2
)

select distinct C1
    , NameList = stuff((select ' ' + fp2.NamesList
        from FirstPass fp2
        where fp.C1 = fp2.C1
        group by fp2.GameCount, fp2.NamesList
        order by fp2.GameCount desc
        FOR XML PATH('')), 1, 1, '')
from FirstPass fp

---

假设现在的值是

 VALUES
('F92', 'Game1', 'b100'),
('F92', 'Game1', 'b200'),
('F92', 'Game2', 'C200'),
('F92', 'Game2', 'D400'),
('F92', 'Game2', 'D500')
;

所以现在游戏2出现了3次。使用肖恩·兰格的代码，结果是按升序排列的，因此游戏1在游戏2之前出现。是否有一种方法可以根据c2列的计数（降序）显示结果，从而使预期结果

F92 | Game2 C200, D400, D500 Game1 b100, b200 

---

扩展这个例子。以下是新的查询：

            CREATE TABLE Table1
            ([C1] varchar(50), [C2] varchar(50), [C3] varchar(50),[C4] varchar(50),[C5] varchar(50) )
            ;

            INSERT INTO Table1
            ([C1], [C2], [C3], [C4], [C5])
            VALUES
            ('F92', 'XBOX','81-94','Game1', 'B350'),
            ('F92', 'XBOX','81-94','Game1', 'B150'),
            ('F92', 'XBOX','76-80','Game2', 'PB100'),
            ('F92', 'XBOX','76-80','Game2', 'PB200'),
            ('F92', 'XBOX','95-97','Game2', 'PB300')
            ;

            with FirstPass as
            (
                SELECT
                 C1, c2, 
                    c3 + ' ' + c4+ + STUFF(
                     (SELECT ', ' +''+ c5
                      FROM Table1
                      WHERE C1 = a.C1 AND C2 = a.C2  AND C3 = a.C3  AND C4 = a.C4  
                      FOR XML PATH (''))
                      , 1, 1, '')  AS NamesList
                , count(*) as GameCount
                FROM Table1 AS a
                GROUP BY C1,C2,c3, c4
            )

            select distinct C1,C2
                , NameList = stuff((select ' ' + fp2.NamesList
                    from FirstPass fp2
                    where fp.C1 = fp2.C1
                    group by fp2.GameCount, fp2.NamesList
                    order by fp2.GameCount
                    FOR XML PATH('')), 1, 1, '')
                    into table2
            from FirstPass fp

            -- concatenate and insert into existing table

            insert into table2 (C1, conc)
            select C1, C2 + ' ' + NameList
            from FirstPass fp

            SELECT *  FROM table2


            drop table table1
            drop table table2

预期结果：

C1  |conc
F92 |XBOX 95-97 GAME2 PB300 76-80 GAME2 PB100, PB200 81-94 GAME1 B350, B150

并将此结果插入现有表“表2”

在这里连接最佳解决方案是否达到预期效果？查询的这一部分让我很反感：

-- concatenate and insert into existing table

            insert into table2 (C1, conc)
            select C1, C2 + ' ' + NameList
            from FirstPass fp

---

发布ddl、样本数据和所需输出是多么好的工作啊。当你努力发布那么多信息时，这真的很容易帮上忙。谢谢就像一种魅力。所以我还有一个问题（不知道我是否应该创建一张新的票）。如果这是可行的，你应该考虑把它标记为答案。