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

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

其他回答

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)
;

这对我有用(SQL Server 2016):

SELECT CarNamesString = STUFF((
         SELECT ',' + [Name]
            FROM tbl_cars
            FOR XML PATH('')
         ), 1, 1, '')

来源如下:https://www.mytecbits.com/

还有一个针对MySQL的解决方案(因为该页面显示在Google for MySQL中):

SELECT [Name],
       GROUP_CONCAT(DISTINCT [Name]  SEPARATOR ',')
       FROM tbl_cars

来自MySQL文档。

PostgreSQL数组非常棒。例子:

创建一些测试数据:

postgres=# \c test
You are now connected to database "test" as user "hgimenez".
test=# create table names (name text);
CREATE TABLE
test=# insert into names (name) values ('Peter'), ('Paul'), ('Mary');
INSERT 0 3
test=# select * from names;
 name
-------
 Peter
 Paul
 Mary
(3 rows)

将它们聚合到一个数组中:

test=# select array_agg(name) from names;
 array_agg
-------------------
 {Peter,Paul,Mary}
(1 row)

将数组转换为逗号分隔的字符串:

test=# select array_to_string(array_agg(name), ', ') from names;
 array_to_string
-------------------
 Peter, Paul, Mary
(1 row)

DONE

由于PostgreSQL 9.0,引用删除的答案“没有名字的马”更容易:

select string_agg(name, ',') 
from names;

对于Oracle DB,请参阅以下问题:如何在Oracle中将多行连接成一行而不创建存储过程?

最好的答案似乎是@Emmanuel,使用了内置的LISTAGG()函数,该函数在Oracle 11g Release 2和更高版本中提供。

SELECT question_id,
   LISTAGG(element_id, ',') WITHIN GROUP (ORDER BY element_id)
FROM YOUR_TABLE;
GROUP BY question_id

正如@user762952所指出的,根据Oracle的文档http://www.oracle-base.com/articles/misc/string-aggregation-techniques.php,WM_CONCAT()函数也是一个选项。它看起来很稳定,但Oracle明确建议不要将它用于任何应用程序SQL,因此使用时风险自负。

除此之外,您还必须编写自己的函数;上面的Oracle文档提供了如何做到这一点的指南。

使用“TABLE”类型非常容易。让我们假设您的表名为Students,并且它具有列名。

declare @rowsCount INT
declare @i INT = 1
declare @names varchar(max) = ''

DECLARE @MyTable TABLE
(
  Id int identity,
  Name varchar(500)
)
insert into @MyTable select name from Students
set @rowsCount = (select COUNT(Id) from @MyTable)

while @i < @rowsCount
begin
 set @names = @names + ', ' + (select name from @MyTable where Id = @i)
 set @i = @i + 1
end
select @names

此示例使用SQL Server 2008 R2进行了测试。