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

在Chris Shaffer的回答之上:

如果您的数据可能会重复,例如

Tom
Ali
John
Ali
Tom
Mike

而不是汤姆、阿里、约翰、阿里、汤姆、迈克

您可以使用DISTINCT避免重复,并让Tom、Ali、John、Mike:

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

其他回答

使用COALESCE-从这里了解更多信息

例如:

102103104

然后在SQL Server中编写以下代码,

Declare @Numbers AS Nvarchar(MAX) -- It must not be MAX if you have few numbers
SELECT  @Numbers = COALESCE(@Numbers + ',', '') + Number
FROM   TableName where Number IS NOT NULL

SELECT @Numbers

输出将是:

102,103,104

SQL Server 2017+和SQL Azure:STRING_AGG

从SQL Server的下一个版本开始,我们终于可以跨行连接,而无需使用任何变量或XML开关。

STRING_AGG(Transact-SQL)

不分组

SELECT STRING_AGG(Name, ', ') AS Departments
FROM HumanResources.Department;

分组时:

SELECT GroupName, STRING_AGG(Name, ', ') AS Departments
FROM HumanResources.Department
GROUP BY GroupName;

带分组和子排序

SELECT GroupName, STRING_AGG(Name, ', ') WITHIN GROUP (ORDER BY Name ASC) AS Departments
FROM HumanResources.Department
GROUP BY GroupName;

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]

如果要处理null,可以通过添加where子句或在第一个子句周围添加另一个COALENCE来完成。

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

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