SQL: psql 中的逻辑问题

SQL: Logic problems in psql

我正在尝试开发一个查询来识别具有多个客户端 ID 的客户端。客户 ID 是第 1 列,第 19 和 20 列包含唯一的个人标识符,您可以将它们视为某种社会安全号码(我们称它们为 SSN.19 和 SSN.20)

我的第一个想法是查找具有匹配 SSN 但客户端 ID 不同的每一行,如下所示:

SELECT  
    a."5", a."3"||' '||a."4" as "3+4", a."19", a."20", a."21", a."1", 
    b."1", a."8"
FROM
    "clients_1" AS a,
    "clients_1" AS b 
WHERE a."19"=b."19" and a."20"=b."20" and a."1"<b."1" and a."1"='Value';

但是,它返回了 0 行。为了检查 table 是否确实没有重复项,我执行了以下查询:

select distinct "19" as hk, count("19") as dl from "clients_1" group by "19" order by dl desc;

select distinct "20" as hk, count("20") as dl from "clients_1" group by "20" order by dl desc;

事实证明,在这个特定的 table 上,没有客户端关联 SSN19,但 table 中有几个重复的 SSN20。所以我执行了以下查询来查找具有多个 ID 的客户:

SELECT  
    a."5", a."3"||' '||a."4" as "3+4", a."20", a."21", a."1", 
    b."1", a."8"
FROM
    "clients_1" AS a,
    "clients_1" AS b 
WHERE a."20"=b."20" and a."1"<b."1" and a."7"='Value';

此人返回了 table,其中包含多个 ID 不同但 SSN20 相同的客户端。之后,我开始考虑一种方法,可以将此查询概括为客户同时拥有 SSN19 和 SSN20 或其中之一的情况,所以我想到了以下几点:

SELECT                
    a."5", a."3"||' '||a."4" as "3+4", a."19", a."20", a."21", a."1", 
    b."1", a."8"
FROM
    "clients_1" AS a,
    "clients_1" AS b 
WHERE ((a."19"=b."19" and a."19" is not null) or (a."20"=b."20" and a."20" is not null)) and a."1"<b."1" and a."7"='Value';

但是,这个查询需要很长时间,我查询 运行 大约 20 分钟,但没有任何结果,而之前的尝试最多用了大约 2 分钟。我究竟做错了什么?

我相信这样的东西会更好地表现并给你更多的灵活性:

SELECT
    *
FROM
    (
        SELECT
            COUNT(*) OVER (PARTITION BY "19") as 19_matches,
            COUNT(*) OVER (PARTITION BY "20") as 20_matches,
            COUNT(*) OVER (PARTITION BY "19","20") as both_matches,
            clients_1.*
        FROM 
            clients_1
        WHERE "7" = 'value'
    )
WHERE 19_matches > 1 OR 20_matches > 1 or both_matches > 1
ORDER BY "19","20"

除了丑陋的列名,这只是WHERE EXISTS(a similar record)解决方案:


SELECT *
FROM clients_1 AS a
WHERE EXISTS(
SELECT* FROM clients_1 AS b 
        WHERE (a."19" = b."19" OR a."20" = b."20" )
        AND a."1" <> b."1" 
        );