Question

我的数据库中有以下关系：

组织：有关政治和经济组织的信息。
名称：组织的全名
缩写：缩写

isMember：政治和经济组织的成员。
组织：组织的缩写
国家：成员国的代码

geo_desert：有关沙漠的地理信息
沙漠：沙漠的名称
国家：所在的国家代码
省：省国家

我的任务是检索在其成员中拥有沙漠的全套国家/地区的组织。这个组织也可以有没有沙漠的国家。所以我有一组有沙漠的国家，结果中的每个组织都应该把所有组织都作为成员和任意数量的其他（没有沙漠）国家。

到目前为止，我尝试编写以下代码，但它无法正常工作。

WITH CountriesWithDeserts AS (
    SELECT DISTINCT country
    FROM dbmaster.geo_desert        
), OrganizationsWithAllDesertMembers AS (
    SELECT organization 
    FROM dbmaster.isMember AS ism 
    WHERE (
        SELECT count(*)
        FROM (
            SELECT *
            FROM  CountriesWithDeserts          
            EXCEPT
            SELECT country
            FROM dbmaster.isMember
            WHERE organization = ism.organization
        )
    ) IS NULL
), OrganizationCode AS (
    SELECT name, abbreviation
    FROM dbmaster.Organization  
)
SELECT oc.name AS Organization
FROM OrganizationCode AS oc, OrganizationsWithAllDesertMembers AS owadm
WHERE oc.abbreviation=owadm.organization;

UPD： DBMS说：＆＃34; ism.organization未定义＆＃34;
我使用的是DB2 / LINUXX8664 9.7.0

输出应如下所示：

NAME
------------------------------------------ --------------------------------------
    非洲，加勒比和太平洋国家
    非洲开发银行     文化技术合作机构
    安第斯集团

Answer 1

我发现处理此问题的最简单方法是使用group by和having。你只想专注于沙漠，所以其他国家都不重要。

select m.organization
from isMember m join
     geo_desert d
     on m.country = d.country
group by m.organization
having count(distinct m.country) = (select count(distinct d.country) from geo_desert);

having条款只计算匹配（即沙漠）国家的数量，并检查所有国家/地区。

Answer 2

这样说：您正在寻找不存在他们不包括的沙漠国家的组织。

select *
from organization o
where not exists
(
  select country from geo_desert
  except
  select country from ismember
  where organization = o.abbreviation
);

Answer 3

以下是两个等效的解决方案：

第一：

WITH CountriesWithDeserts AS (
    SELECT DISTINCT country
    FROM dbmaster.geo_desert        
), OrganizationsWithAllDesertMembers AS (
    SELECT ism.organization 
    FROM dbmaster.isMember AS ism
    JOIN CountriesWithDeserts AS cwd
    ON ism.country = cwd.country
    GROUP BY ism.organization
    HAVING count(ism.country) = (SELECT count(*) FROM CountriesWithDeserts)
), OrganizationCode AS (
    SELECT name, abbreviation
    FROM dbmaster.Organization  
)
SELECT oc.name AS Organization
FROM OrganizationCode AS oc, OrganizationsWithAllDesertMembers AS owadm
WHERE oc.abbreviation=owadm.organization;

第二

WITH CountriesWithDeserts AS (
    SELECT DISTINCT country
    FROM dbmaster.geo_desert        
)
SELECT org.name AS Organization
FROM dbmaster.Organization AS org
WHERE NOT EXISTS (
    SELECT *
    FROM  CountriesWithDeserts          
    EXCEPT
    SELECT country
    FROM dbmaster.isMember
    WHERE organization = org.abbreviation
);

SQL查询将检索包含来自另一个集的所有条目的集合

3 个答案: