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?
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
下面是一个使用“基本循环”和“rownum”实现给定场景的简单PL/SQL过程
表定义
CREATE TABLE "NAMES" ("NAME" VARCHAR2(10 BYTE))) ;
让我们将值插入此表
INSERT INTO NAMES VALUES('PETER');
INSERT INTO NAMES VALUES('PAUL');
INSERT INTO NAMES VALUES('MARY');
程序从这里开始
DECLARE
MAXNUM INTEGER;
CNTR INTEGER := 1;
C_NAME NAMES.NAME%TYPE;
NSTR VARCHAR2(50);
BEGIN
SELECT MAX(ROWNUM) INTO MAXNUM FROM NAMES;
LOOP
SELECT NAME INTO C_NAME FROM
(SELECT ROWNUM RW, NAME FROM NAMES ) P WHERE P.RW = CNTR;
NSTR := NSTR ||','||C_NAME;
CNTR := CNTR + 1;
EXIT WHEN CNTR > MAXNUM;
END LOOP;
dbms_output.put_line(SUBSTR(NSTR,2));
END;
后果
PETER,PAUL,MARY
其他回答
使用此项:
ISNULL(SUBSTRING(REPLACE((select ',' FName as 'data()' from NameList for xml path('')), ' ,',', '), 2, 300), '') 'MyList'
其中“300”可以是任何宽度,考虑到您认为将显示的最大项目数。
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。
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)
;
以下是实现这一目标的完整解决方案:
-- Table Creation
CREATE TABLE Tbl
( CustomerCode VARCHAR(50)
, CustomerName VARCHAR(50)
, Type VARCHAR(50)
,Items VARCHAR(50)
)
insert into Tbl
SELECT 'C0001','Thomas','BREAKFAST','Milk'
union SELECT 'C0001','Thomas','BREAKFAST','Bread'
union SELECT 'C0001','Thomas','BREAKFAST','Egg'
union SELECT 'C0001','Thomas','LUNCH','Rice'
union SELECT 'C0001','Thomas','LUNCH','Fish Curry'
union SELECT 'C0001','Thomas','LUNCH','Lessy'
union SELECT 'C0002','JOSEPH','BREAKFAST','Bread'
union SELECT 'C0002','JOSEPH','BREAKFAST','Jam'
union SELECT 'C0002','JOSEPH','BREAKFAST','Tea'
union SELECT 'C0002','JOSEPH','Supper','Tea'
union SELECT 'C0002','JOSEPH','Brunch','Roti'
-- function creation
GO
CREATE FUNCTION [dbo].[fn_GetItemsByType]
(
@CustomerCode VARCHAR(50)
,@Type VARCHAR(50)
)
RETURNS @ItemType TABLE ( Items VARCHAR(5000) )
AS
BEGIN
INSERT INTO @ItemType(Items)
SELECT STUFF((SELECT distinct ',' + [Items]
FROM Tbl
WHERE CustomerCode = @CustomerCode
AND Type=@Type
FOR XML PATH(''))
,1,1,'') as Items
RETURN
END
GO
-- fianl Query
DECLARE @cols AS NVARCHAR(MAX),
@query AS NVARCHAR(MAX)
select @cols = STUFF((SELECT distinct ',' + QUOTENAME(Type)
from Tbl
FOR XML PATH(''), TYPE
).value('.', 'NVARCHAR(MAX)')
,1,1,'')
set @query = 'SELECT CustomerCode,CustomerName,' + @cols + '
from
(
select
distinct CustomerCode
,CustomerName
,Type
,F.Items
FROM Tbl T
CROSS APPLY [fn_GetItemsByType] (T.CustomerCode,T.Type) F
) x
pivot
(
max(Items)
for Type in (' + @cols + ')
) p '
execute(@query)
此方法仅适用于Teradata Aster数据库,因为它使用NPATH函数。
再次,我们有桌上学生
SubjectID StudentName
---------- -------------
1 Mary
1 John
1 Sam
2 Alaina
2 Edward
然后使用NPATH,只需一次SELECT:
SELECT * FROM npath(
ON Students
PARTITION BY SubjectID
ORDER BY StudentName
MODE(nonoverlapping)
PATTERN('A*')
SYMBOLS(
'true' as A
)
RESULT(
FIRST(SubjectID of A) as SubjectID,
ACCUMULATE(StudentName of A) as StudentName
)
);
结果:
SubjectID StudentName
---------- -------------
1 [John, Mary, Sam]
2 [Alaina, Edward]