SQL Server案例陈述

Question

我正在尝试编写一个查询，每个employerid只生成一行，而不是每个雇主生成多行，并且一次只填充一列。我通过查询获得以下内容：

我正在使用的查询是：

SELECT Employer.employerid,
    CASE WHEN Service.xxserviceid = '1' THEN 'Number 1' ELSE 'NULL' END AS Service1,
    CASE WHEN Service.xxserviceid  = '2' THEN 'Number 2' ELSE 'NULL' END AS Service2,
    CASE WHEN Service.xxserviceid  = '3' THEN 'Number 2' ELSE 'NULL' END AS Service3
FROM Employer
INNER JOIN Service
ON Service.employerid = Employer.employerid;

所以我希望每个雇主在一行中填写Service1，Service2，Service3列，而不是每个雇主多行。

Answer 1

您可以使用GROUP BY + MAX将结果减少到每位员工一行：

SELECT Employer.employerid,
  max(CASE WHEN Service.xxserviceid = '1' THEN 'Number 1' ELSE 'NULL' END) AS Service1,
  max(CASE WHEN Service.xxserviceid  = '2' THEN 'Number 2' ELSE 'NULL' END) AS Service2,
  max(CASE WHEN Service.xxserviceid  = '3' THEN 'Number 3' ELSE 'NULL' END) AS Service3
FROM Employer
INNER JOIN Service
  ON Service.employerid = Employer.employerid
GROUP BY Employer.employerid;

如果你考虑一下，你甚至不需要加入Employer表，因为你所使用的只是employerid列，它已经存在于Service。

因此，查询可以简化为：

SELECT employerid,
  max(CASE WHEN xxserviceid = '1' THEN 'Number 1' ELSE 'NULL' END) AS Service1,
  max(CASE WHEN xxserviceid  = '2' THEN 'Number 2' ELSE 'NULL' END) AS Service2,
  max(CASE WHEN xxserviceid  = '3' THEN 'Number 3' ELSE 'NULL' END) AS Service3
FROM Service
GROUP BY employerid;

SQLFiddle：http://www.sqlfiddle.com/#!6/004a9/2

Answer 2

SELECT Employer.employerid,
    MAX(CASE WHEN Service.xxserviceid = '1' THEN 'Number 1' ELSE 'NULL' END) AS Service1,
    MAX(CASE WHEN Service.xxserviceid  = '2' THEN 'Number 2' ELSE 'NULL' END) AS Service2,
    MAX(CASE WHEN Service.xxserviceid  = '3' THEN 'Number 2' ELSE 'NULL' END) AS Service3
FROM Employer
INNER JOIN Service
ON Service.employerid = Employer.employerid;
GROUP BY Employer.employerID

Answer 3

据我了解您的问题，您可能只需要一个服务列。 这篇文章中提供的其他答案应该有效。如果您只需要一个“服务”列，则可以尝试以下操作：

SELECT Employer.employerid,
    'Number ' + Max(Service.xxserviceid) AS Service1
FROM Employer
INNER JOIN Service
ON Service.employerid = Employer.employerid;
GROUP BY Employer.employerid