考虑一个包含名称的数据库表,其中有三行:

Peter
Paul
Mary

有没有一种简单的方法可以把它变成彼得、保罗、玛丽的一串?


当前回答

提出了递归CTE解决方案,但没有提供代码。下面的代码是递归CTE的示例。

请注意,虽然结果与问题相符,但数据与给定的描述并不完全相符,因为我假设您确实希望对行组(而不是表中的所有行)执行此操作。将其更改为与表中的所有行相匹配是读者的练习。

;WITH basetable AS (
    SELECT
        id,
        CAST(name AS VARCHAR(MAX)) name,
        ROW_NUMBER() OVER (Partition BY id ORDER BY seq) rw,
        COUNT(*) OVER (Partition BY id) recs
    FROM (VALUES
        (1, 'Johnny', 1),
        (1, 'M', 2),
        (2, 'Bill', 1),
        (2, 'S.', 4),
        (2, 'Preston', 5),
        (2, 'Esq.', 6),
        (3, 'Ted', 1),
        (3, 'Theodore', 2),
        (3, 'Logan', 3),
        (4, 'Peter', 1),
        (4, 'Paul', 2),
        (4, 'Mary', 3)
    ) g (id, name, seq)
),
rCTE AS (
    SELECT recs, id, name, rw
    FROM basetable
    WHERE rw = 1

    UNION ALL

    SELECT b.recs, r.ID, r.name +', '+ b.name name, r.rw + 1
    FROM basetable b
    INNER JOIN rCTE r ON b.id = r.id AND b.rw = r.rw + 1
)
SELECT name
FROM rCTE
WHERE recs = rw AND ID=4
OPTION (MAXRECURSION 101)

其他回答

提出了递归CTE解决方案,但没有提供代码。下面的代码是递归CTE的示例。

请注意,虽然结果与问题相符,但数据与给定的描述并不完全相符,因为我假设您确实希望对行组(而不是表中的所有行)执行此操作。将其更改为与表中的所有行相匹配是读者的练习。

;WITH basetable AS (
    SELECT
        id,
        CAST(name AS VARCHAR(MAX)) name,
        ROW_NUMBER() OVER (Partition BY id ORDER BY seq) rw,
        COUNT(*) OVER (Partition BY id) recs
    FROM (VALUES
        (1, 'Johnny', 1),
        (1, 'M', 2),
        (2, 'Bill', 1),
        (2, 'S.', 4),
        (2, 'Preston', 5),
        (2, 'Esq.', 6),
        (3, 'Ted', 1),
        (3, 'Theodore', 2),
        (3, 'Logan', 3),
        (4, 'Peter', 1),
        (4, 'Paul', 2),
        (4, 'Mary', 3)
    ) g (id, name, seq)
),
rCTE AS (
    SELECT recs, id, name, rw
    FROM basetable
    WHERE rw = 1

    UNION ALL

    SELECT b.recs, r.ID, r.name +', '+ b.name name, r.rw + 1
    FROM basetable b
    INNER JOIN rCTE r ON b.id = r.id AND b.rw = r.rw + 1
)
SELECT name
FROM rCTE
WHERE recs = rw AND ID=4
OPTION (MAXRECURSION 101)

我通常使用这样的select连接SQL Server中的字符串:

with lines as 
( 
  select 
    row_number() over(order by id) id, -- id is a line id
    line -- line of text.
  from
    source -- line source
), 
result_lines as 
( 
  select 
    id, 
    cast(line as nvarchar(max)) line 
  from 
    lines 
  where 
    id = 1 
  union all 
  select 
    l.id, 
    cast(r.line + N', ' + l.line as nvarchar(max))
  from 
    lines l 
    inner join 
    result_lines r 
    on 
      l.id = r.id + 1 
) 
select top 1 
  line
from
  result_lines
order by
  id desc

此答案可能会返回意外的结果。要获得一致的结果,请使用其他答案中详细说明的For XML PATH方法之一。

使用COALENCE:

DECLARE @Names VARCHAR(8000) 
SELECT @Names = COALESCE(@Names + ', ', '') + Name 
FROM People

只是一些解释(因为这个答案似乎得到了相对规律的观点):

联合实际上只是一种有助于实现两件事的欺骗:

1) 无需使用空字符串值初始化@Names。

2) 无需在末端去除额外的分隔符。

如果一行具有NULL Name值(如果存在NULL,NULL将使该行之后的@Names为NULL,而下一行将再次以空字符串开始),则上述解决方案将给出错误的结果。使用以下两种解决方案之一即可轻松解决:

DECLARE @Names VARCHAR(8000) 
SELECT @Names = COALESCE(@Names + ', ', '') + Name
FROM People
WHERE Name IS NOT NULL

or:

DECLARE @Names VARCHAR(8000) 
SELECT @Names = COALESCE(@Names + ', ', '') + 
    ISNULL(Name, 'N/A')
FROM People

取决于您想要的行为(第一个选项只是过滤掉NULL,第二个选项用标记消息将它们保留在列表中[用适合您的内容替换“N/a”])。

SQL Server 2005或更高版本

CREATE TABLE dbo.Students
(
    StudentId INT
    , Name VARCHAR(50)
    , CONSTRAINT PK_Students PRIMARY KEY (StudentId)
);

CREATE TABLE dbo.Subjects
(
    SubjectId INT
    , Name VARCHAR(50)
    , CONSTRAINT PK_Subjects PRIMARY KEY (SubjectId)
);

CREATE TABLE dbo.Schedules
(
    StudentId INT
    , SubjectId INT
    , CONSTRAINT PK__Schedule PRIMARY KEY (StudentId, SubjectId)
    , CONSTRAINT FK_Schedule_Students FOREIGN KEY (StudentId) REFERENCES dbo.Students (StudentId)
    , CONSTRAINT FK_Schedule_Subjects FOREIGN KEY (SubjectId) REFERENCES dbo.Subjects (SubjectId)
);

INSERT dbo.Students (StudentId, Name) VALUES
    (1, 'Mary')
    , (2, 'John')
    , (3, 'Sam')
    , (4, 'Alaina')
    , (5, 'Edward')
;

INSERT dbo.Subjects (SubjectId, Name) VALUES
    (1, 'Physics')
    , (2, 'Geography')
    , (3, 'French')
    , (4, 'Gymnastics')
;

INSERT dbo.Schedules (StudentId, SubjectId) VALUES
    (1, 1)        --Mary, Physics
    , (2, 1)    --John, Physics
    , (3, 1)    --Sam, Physics
    , (4, 2)    --Alaina, Geography
    , (5, 2)    --Edward, Geography
;

SELECT
    sub.SubjectId
    , sub.Name AS [SubjectName]
    , ISNULL( x.Students, '') AS Students
FROM
    dbo.Subjects sub
    OUTER APPLY
    (
        SELECT
            CASE ROW_NUMBER() OVER (ORDER BY stu.Name) WHEN 1 THEN '' ELSE ', ' END
            + stu.Name
        FROM
            dbo.Students stu
            INNER JOIN dbo.Schedules sch
                ON stu.StudentId = sch.StudentId
        WHERE
            sch.SubjectId = sub.SubjectId
        ORDER BY
            stu.Name
        FOR XML PATH('')
    ) x (Students)
;

在SQL Server 2005中

SELECT Stuff(
  (SELECT N', ' + Name FROM Names FOR XML PATH(''),TYPE)
  .value('text()[1]','nvarchar(max)'),1,2,N'')

在SQL Server 2016中

可以使用FOR JSON语法

SELECT per.ID,
Emails = JSON_VALUE(
   REPLACE(
     (SELECT _ = em.Email FROM Email em WHERE em.Person = per.ID FOR JSON PATH)
    ,'"},{"_":"',', '),'$[0]._'
) 
FROM Person per

结果会变成

Id  Emails
1   abc@gmail.com
2   NULL
3   def@gmail.com, xyz@gmail.com

即使您的数据包含无效的XML字符,这也会起作用

“”},{“_”:“”是安全的,因为如果您的数据包含“”},{”_“:“”,它将被转义为“},{\”_\“:\”

可以用任何字符串分隔符替换“,”


在SQL Server 2017中,Azure SQL数据库

您可以使用新的STRING_AGG函数