Question

我正在处理人口普查数据，我希望通过提供名单列表来提供搜索记录的功能。这个想法是，如果您知道名称或2或3个家庭成员，您将能够排除所有没有这些名称的地址。请考虑此示例数据集（地址，姓氏，名称）：

“在Janemount Lower（Cork No. 4 Urban（部分），Cork）的房子2的居民”，“赌博”，“Julia”
“Janemount Lower（Cork No. 4 Urban（部分），科克）的房子2的居民”，“赌博”，“理查德”
“在Janemount Lower（Cork No. 4 Urban（部分），Cork）的房子2的居民”，“赌博”，“Hannah”
“住在Janemount Lower（Cork No. 4 Urban（部分），科克）的房子2”，“赌博”，“海伦”

搜索Julia，Hannah和Helen应该能够返回所有4行，因为它们共享一个公共地址。这听起来很简单，但我遇到了这个问题。由于数据集的大小，游标已经不存在了。有什么想法吗？

（不用说，我已经简化了这一点，因为我现在忽略了搜索中的姓氏部分）

Answer 1

你可以试试这个：

SELECT A.*
FROM YourTable A
JOIN (  SELECT Address, COUNT(*) Quant
        FROM YourTable
        WHERE Forename IN ('Julia','Hannah','Helen')
        GROUP BY Address
        HAVING COUNT(DISTINCT Forename) > 2) B
ON A.Address = B.Address

Answer 2

这是relational divisio n问题。

SELECT Address, FamilyName, Forename
FROM   YourTable
WHERE  Address IN (SELECT Address
                   FROM   YourTable
                   WHERE  Forename IN ( 'Julia', 'Hannah', 'Helen' )
                   GROUP  BY Address
                   HAVING COUNT(DISTINCT Forename) = 3)

或者

WITH Names(name)
     AS (SELECT 'Julia'
         UNION ALL
         SELECT 'Hannah'
         UNION ALL
         SELECT 'Helen')
SELECT Address,
       FamilyName,
       Forename
FROM   YourTable y1
WHERE  NOT EXISTS (SELECT *
                   FROM   Names n
                   WHERE  NOT EXISTS(SELECT *
                                     FROM   YourTable y2
                                     WHERE  y1.Address = y2.Address
                                            AND y2.Forename = n.Name))

Answer 3

此查询：

       select streetaddress, count(*) as occupantcount
       from census
       where firstname in ("Julia", "Hannah", "Helen")
       group by streetaddress
       order by occupantcount desc

将返回每个地址中占用者的地址和数量，只要其中一个占用者具有IN（）列表中的一个名字，并按最大占用者到最少占用者顺序排序结果。对同一原则的变化（按地址分组）可以为您提供其他类型的信息。您可以对结果应用连续的过滤器，以便在所需的行上进行调零。

Answer 4

SELECT t.address
     , t.familyname
     , t.forename 
  FROM yourTable t
  WHERE t.address IN 
    -- search subquery
    ( SELECT s1.address 
        FROM yourTable s1
          JOIN yourTable s2
            ON s2.address = s1.address
          JOIN yourTable s3
            ON s3.address = s1.address
        WHERE s1.forename = "Julia"
          AND s2.forename = "Hannah"
          AND s3.forename = "Helen"
    )
  ORDER BY t.address
         , t.familyname
         , t.forename 
;

第二个解决方案：

SELECT t.address
     , t.familyname
     , t.forename 
  FROM yourTable t
  WHERE EXISTS 
    -- search subquery
    ( SELECT * 
        FROM yourTable s1
          JOIN yourTable s2
            ON s2.address = s1.address
          JOIN yourTable s3
            ON s3.address = s1.address
        WHERE s1.forename = "Julia"
          AND s2.forename = "Hannah"
          AND s3.forename = "Helen"
          AND s1.address = t.address
    )
  ORDER BY t.address
         , t.familyname
         , t.forename 
;

第3个解决方案：

SELECT t.address
     , t.familyname
     , t.forename
  FROM yourTable t
  WHERE  -- search subqueries
    EXISTS
      ( SELECT * 
          FROM yourTable s1
            WHERE s1.forename= "Julia"
              AND s1.address = t.address
      )
    AND EXISTS
      ( SELECT * 
          FROM yourTable s2
            WHERE s2.forename = "Hannah"
              AND s2.address = t.address
      )
    AND EXISTS
      ( SELECT * 
          FROM yourTable s3
            WHERE s3.forename = "Helen"
              AND s3.address = t.address
      )
  ORDER BY t.address
         , t.familyname
         , t.forename 
;

Answer 5

如果你索引forename和地址，这应该是直截了当的：

SELECT a.Address as CommonAddress
FROM (SELECT Address FROM Names WHERE Forename = 'Julia') a
INNER JOIN (SELECT Address FROM Names WHERE Forename = 'Richard') b ON a.Address=b.Address
INNER JOIN (SELECT Address FROM Names WHERE Forename = 'Helen') c on b.Address=c.Address

Answer 6

解决方案是使用子查询查找所有提供的名称共有的地址，并返回该地址的所有记录。

declare @people table (
    address varchar(255),
    familyName varchar(255),
    forename varchar(255)
)

insert into @people
values ('Residents of a house 2 in Janemount Lower (Cork No. 4 Urban (part of), Cork)', 'Gamble', 'Julia')
insert into @people
values ('Residents of a house 2 in Janemount Lower (Cork No. 4 Urban (part of), Cork)', 'Gamble', 'Richard')
insert into @people
values ('Residents of a house 2 in Janemount Lower (Cork No. 4 Urban (part of), Cork)', 'Gamble', 'Hannah')
insert into @people
values ('Residents of a house 2 in Janemount Lower (Cork No. 4 Urban (part of), Cork)', 'Gamble', 'Helen')
insert into @people
values ('Residents of a house 2 somewhere else (Cork No. 4 Urban (part of), Cork)', 'Cooper', 'Helen')

select people.*
from @people as people
where people.address in (
    select address
    from @people
    where forename in ('Julia', 'Hannah', 'Helen')
    group by address
    having count(forename) >= 3 -- This must be equal to the number of names searched for
)

这个解决方案的一个问题是，如果有三个汉娜住在同一个地址，那么即使朱莉娅和海伦不住在那里，查询也会归还那些人。

Answer 7

DECLARE @namecount int;
DECLARE @forenames TABLE (name varchar(50));
INSERT INTO @forenames
VALUES ('...'),
       ('...'),
       ('...');
SELECT @namecount = COUNT(*) FROM @forenames;

/* list all people by addresses that are shared by people
   whose forenames are included in @forenames */
SELECT cd.*
FROM CensusData cd
  INNER JOIN (
    SELECT d.Address
    FROM CensusData d
      INNER JOIN @forenames f ON d.Forename = f.Name
    GROUP BY d.Address
    HAVING COUNT(DISTINCT d.Forename) >= @namecount
  ) filter ON cd.Address = filter.Address

/* same for family names */
SELECT cd.*
FROM CensusData cd
  INNER JOIN (
    SELECT d.[Family Name]
    FROM CensusData d
      INNER JOIN @forenames f ON d.Forename = f.Name
    GROUP BY d.Address
    HAVING COUNT(DISTINCT d.Forename) >= @namecount
  ) filter ON cd.[Family Name]= filter.[Family Name]

/* and so on fro other criteria */

如果需要，您还可以组合标准。

Answer 8

select *
from dataset
where address = (
    select address
    from dataset
    where famly_name = 'Hannah' and forename = 'Gamble'
)

Answer 9

你可以从这样的事情开始，假设地址要完全匹配。它需要调整，因为你很可能会看到一些重复

Select
    T1.*
From
    TableName T1
Inner Join
    TableName T2
On
    T1.Address = T2.Address
Where
    T1.ForeName = 'Julia'

SQL查询问题 - 如何使用共性选择行组

9 个答案: