[英]SQL Select First, Second, Third, Fourth ,Last
我有一个会员数据库,其中有居住在同一地址的人数。 为了最大程度减少邮寄新闻通讯时的邮政支出,我想从数据库中选择唯一的地址,并将所有姓名添加在居住在同一地址的人的同一标签上。 我在Members
表中具有的字段是Names
, AddressLine1
, AddressLine2
, Suburb
, State
, Postcode
。
我很高兴创建一个新表,例如,如果需要, LastOfNames
可以使用其他名为FirstOfName
, SecondOfNames
, ThirdOfNames
, LastOfNames
字段。 我的数据库中居住在同一地址的最大人数为5。
标签上要求的输出示例:
John Smith
Jane Smith
10 High Street
Beverly Hills, NSW 2000
非常感谢您的帮助,因为我对此有很多困难。
感谢堆,米卡
我认为您的理想目标是在一个复杂的查询中最多包含五个名称和一个地址。 该示例在PostgreSQL中。 您没有指定数据库,但我相信ORACLE和SQLServer中存在推论。 您将必须进行自己的改编。
它有点混乱,但可以。 简而言之,您可以使用WINDOW函数对每个唯一地址的多个名称进行排名。 使用WITH子句按等级隔离每个名称。 左联接为少于五个名称的地址选择名称。
样表:
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;
这是最终的输出:
“简”;“史密斯”;“约翰”;“史密斯”;“”;“”“”“”“”“”;“”;“ 10高街”;“新南威尔士州比佛利山庄2000”;“ CA” ;“ 90210”“ Jack”;“ Smith”;“ Jean”;“ Smith”;“ Jeff”;“ Smith”;“ Jenn”;“ Smith”;“ Josh”;“ Smith”;“ 12 High Street”; “比佛利山庄NSW 2000”;“ CA”;“ 90210”
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