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SQL选择第一、第二、第三、第四、最后_Sql_Database_Select_Group By - Fatal编程技术网

SQL选择第一、第二、第三、第四、最后

SQL选择第一、第二、第三、第四、最后,sql,database,select,group-by,Sql,Database,Select,Group By,我有一个会员数据库,里面有许多住在同一地址的人。为了在邮寄时事通讯时尽可能减少邮寄费用,我想从数据库中选择唯一的地址,并将住在同一地址的人的所有姓名添加到同一标签标签上。我在成员表中的字段有姓名,地址行1,地址行2,郊区,州,邮政编码 我很乐意创建一个新表,例如,如果需要,可以使用名为FirstOfName,SecondOfNames,ThirdOfNames,LastOfNames的附加字段。在我的数据库中,居住在同一地址的人数最多为5人 标签上所需输出的示例: John Smith Jan

我有一个会员数据库,里面有许多住在同一地址的人。为了在邮寄时事通讯时尽可能减少邮寄费用,我想从数据库中选择唯一的地址,并将住在同一地址的人的所有姓名添加到同一标签标签上。我在
成员
表中的字段有
姓名
地址行1
地址行2
郊区
邮政编码

我很乐意创建一个新表,例如,如果需要,可以使用名为
FirstOfName
SecondOfNames
ThirdOfNames
LastOfNames
的附加字段。在我的数据库中,居住在同一地址的人数最多为5人

标签上所需输出的示例:

John Smith

Jane Smith

10 High Street

Beverly Hills, NSW 2000
非常感谢您的帮助,因为我在这方面遇到了很多困难

谢谢你,好几堆,
Mika

我假设您的理想目标是在一个复杂的查询中获得最多五个姓名和一个地址。这个例子在PostgreSQL中。您没有指定数据库,但我相信ORACLE和SQLServer中存在相应的结果。你必须做出自己的调整

虽然有点凌乱,但效果不错。简而言之,您可以使用窗口函数对每个唯一地址的多个名称进行排序。使用WITH子句按等级隔离每个名称。LEFT JOIN可为少于五个名称的地址选择名称

样本表:

   DROP TABLE name_address;

   CREATE TABLE name_address
   (
   first_name CHARACTER VARYING (50)
   ,last_name CHARACTER VARYING (50)
   ,addr_line1 CHARACTER VARYING (50)
   ,city CHARACTER VARYING (50)
   ,state CHARACTER VARYING (2)
   ,zip  CHARACTER VARYING (5)
   )
   WITH (OIDS=TRUE);
   ALTER TABLE name_address OWNER TO your_name_here;
样本数据:

   INSERT INTO name_address (first_name, last_name, addr_line1, city, state, zip) SELECT 'John', 'Smith', '10 High Street', 'Beverly Hills NSW 2000', 'CA', '90210';
   INSERT INTO name_address (first_name, last_name, addr_line1, city, state, zip) SELECT 'Jane', 'Smith', '10 High Street', 'Beverly Hills NSW 2000', 'CA', '90210';
   INSERT INTO name_address (first_name, last_name, addr_line1, city, state, zip) SELECT 'Jeff', 'Smith', '12 High Street', 'Beverly Hills NSW 2000', 'CA', '90210';
   INSERT INTO name_address (first_name, last_name, addr_line1, city, state, zip) SELECT 'Jean', 'Smith', '12 High Street', 'Beverly Hills NSW 2000', 'CA', '90210';
   INSERT INTO name_address (first_name, last_name, addr_line1, city, state, zip) SELECT 'Jenn', 'Smith', '12 High Street', 'Beverly Hills NSW 2000', 'CA', '90210';
   INSERT INTO name_address (first_name, last_name, addr_line1, city, state, zip) SELECT 'Jack', 'Smith', '12 High Street', 'Beverly Hills NSW 2000', 'CA', '90210';
   INSERT INTO name_address (first_name, last_name, addr_line1, city, state, zip) SELECT 'Josh', 'Smith', '12 High Street', 'Beverly Hills NSW 2000', 'CA', '90210';
解决方案:

   WITH nr AS (SELECT DISTINCT first_name, last_name, addr_line1, city, state, zip, COUNT(*) OVER(PARTITION BY zip, state, city, addr_line1 ORDER BY zip, state, city, addr_line1, last_name, first_name) AS name_rank
   FROM name_address
   GROUP BY first_name, last_name, addr_line1, city, state, zip
   ORDER BY zip, state, city, addr_line1, name_rank)
   ,n1 AS(SELECT nr.first_name, nr.last_name, nr.addr_line1, nr.city, nr.state, nr.zip
   FROM nr
   WHERE nr.name_rank = 1)
   ,n2 AS(SELECT nr.first_name, nr.last_name, nr.addr_line1, nr.city, nr.state, nr.zip
   FROM nr
   WHERE nr.name_rank = 2)
   ,n3 AS(SELECT nr.first_name, nr.last_name, nr.addr_line1, nr.city, nr.state, nr.zip
   FROM nr
   WHERE nr.name_rank = 3)
   ,n4 AS(SELECT nr.first_name, nr.last_name, nr.addr_line1, nr.city, nr.state, nr.zip
   FROM nr
   WHERE nr.name_rank = 4)
   ,n5 AS(SELECT nr.first_name, nr.last_name, nr.addr_line1, nr.city, nr.state, nr.zip
   FROM nr
   WHERE nr.name_rank = 5)
   SELECT   DISTINCT n1.first_name
       , n1.last_name
       , n2.first_name
       , n2.last_name
       , n3.first_name
       , n3.last_name
       , n4.first_name
       , n4.last_name
       , n5.first_name
       , n5.last_name
       , na.addr_line1
       , na.city
       , na.state
       , na.zip
   FROM name_address AS na
   LEFT JOIN n1 ON n1.addr_line1 = na.addr_line1 AND n1.city = na.city AND n1.state = na.state AND n1.zip = na.zip
   LEFT JOIN n2 ON n2.addr_line1 = na.addr_line1 AND n2.city = na.city AND n2.state = na.state AND n2.zip = na.zip
   LEFT JOIN n3 ON n3.addr_line1 = na.addr_line1 AND n3.city = na.city AND n3.state = na.state AND n3.zip = na.zip
   LEFT JOIN n4 ON n4.addr_line1 = na.addr_line1 AND n4.city = na.city AND n4.state = na.state AND n4.zip = na.zip
   LEFT JOIN n5 ON n5.addr_line1 = na.addr_line1 AND n5.city = na.city AND n5.state = na.state AND n5.zip = na.zip;
以下是最终输出:

“简”;“史密斯”;“约翰”;“史密斯”;"";"";"";"";"";"";"高街十号;;“新南威尔士州贝弗利山庄2000”;“CA”;"90210" “杰克”;“史密斯”;“Jean”;“史密斯”;“杰夫”;“史密斯”;“詹”;“史密斯”;“乔希”;“史密斯”;“高街12号”;“新南威尔士州贝弗利山庄2000”;“CA”;“90210”