Sql server 将行数据转换为动态列

我知道以前也有人问过类似的问题，但我深陷困境，需要帮助。我有一个类似的数据集

FIRSTNAME  LastNAME  BCODE  Boption           ISDEPENDENT  CRELATIONSHIP  CLASTNAME  CFIRSTNAME  BadgeNum 
BigBird    Street      VIS    Employee+Family  Y           Child          Bert        Street     1234
BigBird    Street      VIS    Employee+Family  Y           Child          Ernie       Street     1234
Oscar      TheGrouch   VIS    EmployeeOnly     N           Null           Null        Null       9090

我需要在SQL中像这样转换数据。依赖列应该是动态的

FIRSTNAME  LASTNAME  BCODE  Dependent 1LN  Dependent 1FN  Dependent 2LN Dependent 2FN  
BigBird    Street     VIS    Street         Bert          Street        Ernie   
Oscar      TheGrouch  VIS     null          null          Null          Null       

---

任何帮助都会很好。

如果不使用动态SQL，没有很多简单的方法可以满足您的要求；但是，这里有一种方法可以尝试

注意：我对您的数据做了一些假设，即badgenum是一个将家属与员工联系起来的值

/* Mock-up table and data */

DECLARE @Data table (
    firstname varchar(50), lastname varchar(50), bcode varchar(3), boption varchar(50), isdependent varchar(1), crelationship varchar(10), clastname varchar(50), cfirstname varchar(50), badgenum int
);

INSERT INTO @Data (
    firstname, lastname, bcode, boption, isdependent, crelationship, cfirstname, clastname, badgenum
) VALUES
    ( 'BigBird', 'Street', 'VIS', 'Employee+Family', 'Y', 'Child', 'Bert', 'Street', 1234 ),
    ( 'BigBird', 'Street', 'VIS', 'Employee+Family', 'Y', 'Child', 'Ernie', 'Street', 1234 ),
    ( 'Oscar', 'TheGrouch', 'VIS', 'EmployeeOnly', 'N', NULL, NULL, NULL, 9090 );

/* Select employees with dependents listed linearly */

;WITH emp AS (
    SELECT DISTINCT
        firstname, lastname, bcode, badgenum
    FROM @Data
)
SELECT
    firstname,
    lastname,
    bcode,
    dep.dep_xml.value( 'data(//dependents/dependent[@id="1"]/lname)[1]', 'varchar(50)' ) AS [Dependent 1LN],
    dep.dep_xml.value( 'data(//dependents/dependent[@id="1"]/fname)[1]', 'varchar(50)' ) AS [Dependent 1FN],
    dep.dep_xml.value( 'data(//dependents/dependent[@id="2"]/lname)[1]', 'varchar(50)' ) AS [Dependent 2LN],
    dep.dep_xml.value( 'data(//dependents/dependent[@id="2"]/fname)[1]', 'varchar(50)' ) AS [Dependent 2FN],
    dep.dep_xml.value( 'data(//dependents/dependent[@id="3"]/lname)[1]', 'varchar(50)' ) AS [Dependent 3LN],
    dep.dep_xml.value( 'data(//dependents/dependent[@id="3"]/fname)[1]', 'varchar(50)' ) AS [Dependent 3FN]
    -- etc...
FROM emp
OUTER APPLY (

    SELECT CAST ( ( 
        SELECT
            ROW_NUMBER() OVER ( ORDER BY d.firstname, d.lastname ) AS '@id',
            d.cfirstname AS fname,
            d.clastname AS lname
        FROM @Data d
        WHERE
            d.badgenum = emp.badgenum
            AND d.isdependent = 'Y'
        FOR XML PATH ( 'dependent' ), ROOT ( 'dependents' )
    ) AS xml ) AS dep_xml

) AS dep;

返回

+-----------+-----------+-------+---------------+---------------+---------------+---------------+---------------+---------------+
| firstname | lastname  | bcode | Dependent 1LN | Dependent 1FN | Dependent 2LN | Dependent 2FN | Dependent 3LN | Dependent 3FN |
+-----------+-----------+-------+---------------+---------------+---------------+---------------+---------------+---------------+
| BigBird   | Street    | VIS   | Street        | Bert          | Street        | Ernie         | NULL          | NULL          |
| Oscar     | TheGrouch | VIS   | NULL          | NULL          | NULL          | NULL          | NULL          | NULL          |
+-----------+-----------+-------+---------------+---------------+---------------+---------------+---------------+---------------+

---

到目前为止，您尝试过什么，或者您所看到的您不理解的问题（以及这些问题是什么）？您好，这里是一个我用解决方案查看的问题，但没有帮助。这是因为列名不在数据集中，因此依赖项的名字和姓氏必须是列名，这不是我需要的。我希望这能回答你的问题。非常感谢@不客气。如果这是针对您的询问，请接受答复。