Question

我知道这可能听起来有些混乱，但基本上我有一个查询返回一组＆＃34;群集名称＆＃34;和＆＃34; ID＆＃34;。我希望查询只为每个ID选择一行，基于＆＃34;群集名称＆＃34;，按字母顺序排序。

例如，我们当前正在让查询返回以下内容：

Cluster Name                                    ID
Construction Technologies          239378769
Construction Technologies          239378942
Construction Technologies          239510698
Law and Public Safety              239510698
Health Science                     239510698
Health Science                     240236166
Health Science                     240236203
Construction Technologies          240236209
Health Science                     240236236
Education and Training             240236303

如果您注意到ID 239510698出现三次，建筑技术，法律和公共安全以及健康科学。我们想要的不是将这行返回三次，而是只返回一次，这将是按字母顺序排列的第一行，即Construction Technologies。

以下是查询：

SELECT c.clustername AS ClusterName, MAX(cc.monsterid) AS CountOfClusters
              FROM tblCareerCluster cc
              INNER JOIN tblClusters c ON c.clusterid = cc.clusterid
              LEFT JOIN tblStudentPersonal sp ON sp.monsterid = cc.monsterid
              INNER JOIN tblStudentSchool ss ON ss.monsterid = cc.monsterid
              INNER JOIN tblSchools s ON s.schoolid = ss.schoolid
              INNER JOIN tblSchoolDistricts sd ON sd.schoolid = s.schoolid
              INNER JOIN tblDistricts d ON d.districtid = sd.districtid
              INNER JOIN tblDistrictUserDistrictGroups rurg ON rurg.schoolid = ss.schoolid
              INNER JOIN tblGroups g ON g.groupid = rurg.groupid
              INNER JOIN tblUserGroups ug ON ug.groupid = g.groupid
              WHERE cc.ranking = (SELECT MAX(cc2.ranking) from tblCareerCluster cc2 INNER JOIN tblCareerCluster cc ON cc.monsterid = cc2.monsterid)  
              AND ss.graduationyear IN (SELECT Items FROM FN_Split('2015,2016,2017,2018,2019,2020,2021,2022,2023', ',')) AND sp.optin = 'Yes'
              AND g.groupname = 'My Districts' AND ug.userid = 14                       
              GROUP BY c.clustername, cc.ranking,  d.district, cc.monsterid, cc.clusterid, d.IRN, d.districtid

如果您需要任何其他信息，请告诉我们，我们将非常感谢您的帮助。

Answer 1

select MIN(A.ClusterName), A.CountOfClusters
FROM
(
    SELECT c.clustername AS ClusterName, MAX(cc.monsterid) AS CountOfClusters
                  FROM tblCareerCluster cc
                  INNER JOIN tblClusters c ON c.clusterid = cc.clusterid
                  LEFT JOIN tblStudentPersonal sp ON sp.monsterid = cc.monsterid
                  INNER JOIN tblStudentSchool ss ON ss.monsterid = cc.monsterid
                  INNER JOIN tblSchools s ON s.schoolid = ss.schoolid
                  INNER JOIN tblSchoolDistricts sd ON sd.schoolid = s.schoolid
                  INNER JOIN tblDistricts d ON d.districtid = sd.districtid
                  INNER JOIN tblDistrictUserDistrictGroups rurg ON rurg.schoolid = ss.schoolid
                  INNER JOIN tblGroups g ON g.groupid = rurg.groupid
                  INNER JOIN tblUserGroups ug ON ug.groupid = g.groupid
                  WHERE cc.ranking = (SELECT MAX(cc2.ranking) from tblCareerCluster cc2 INNER JOIN tblCareerCluster cc ON cc.monsterid = cc2.monsterid)  
                  AND ss.graduationyear IN (SELECT Items FROM FN_Split('2015,2016,2017,2018,2019,2020,2021,2022,2023', ',')) AND sp.optin = 'Yes'
                  AND g.groupname = 'My Districts' AND ug.userid = 14                       
                  GROUP BY c.clustername, cc.ranking,  d.district, cc.monsterid, cc.clusterid, d.IRN, d.districtid
) A
GROUP BY A.CountOfClusters

Answer 2

我认为这个可能更简单

使用返回Cluster Name和ID

的第一个查询

SQL Fiddle Demo

 SELECT MIN(Cluster_Name) as ClusterName, ID
 FROM <your query>
 GROUP BY ID

Answer 3

使用Row_Number为ClusterName排序的每个ID选择第一个结果：

With CTE as (select *, Row_Number() over (partition by ID order by ClusterName) as RN from [Long sequence of tables and joins])

Select ClusterName, ID from CTE where RN = 1

如果您愿意，可以将CTE转换为临时表或子查询。

Answer 4

您可以使用row_number窗口函数对ID的每个分区中的行进行编号，然后选择行号= 1.如果无法将函数直接添加到查询中，则可以使用公用表表达式并将原始查询用作派生表：

with cte as (
  select *, rn = row_number() over (partition by id order by clustername) 
  from ( << your_original_query here >> ) t
)

select * from cte
where rn = 1
order by ClusterName

运行查询，返回按另一列

4 个答案: