Question

架构

CREATE TABLE person
(
     [first_name] VARCHAR(10), 
     [surname] VARCHAR(10), 
     [dob] DATE, 
     [person_id] INT
);

INSERT INTO person ([first_name], [surname], [dob] ,[person_id]) 
VALUES
    ('Alice', 'AA', '1/1/1960', 1),
    ('Bob'  , 'AA', '1/1/1980', 2),
    ('Carol', 'AA', '1/1/2018', 3),
    ('Dave' , 'BB', '1/1/1960', 4),
    ('Elsa', ' BB', '1/1/1980', 5),
    ('Fred' , 'BB', '1/1/1990', 6),
    ('Gina' , 'BB', '1/1/2018', 7);

CREATE TABLE person_membership
(
    [person_id] INT, 
    [personstatus] VARCHAR(1), 
    [membership_id] INT, 
    [relationship] INT
);

INSERT INTO person_membership ([person_id], [personstatus], [membership_id], [relationship])
VALUES
    (1, 'A', 10, 1),
    (2, 'A', 10, 2),
    (3, 'A', 10, 3),
    (4, 'A', 20, 1),
    (5, 'A', 20, 2),
    (6, 'A', 20, 4),
    (7, 'A', 20, 5);

在此简化方案中，关系设置为1的人是主要保单持有人，而不同的数字显示其他人与主要保单持有人的关系（配偶，孩子等）

问题

显示每个主要保单持有人的所有受抚养人，并将其分组为任意选择的年龄组。

所需的输出：

person_id|membership_id|first_name|dependants under 10|dependants over 10
---------+-------------+----------+-------------------+-------------------
       1 |          10 |   Alice  |               Bob |            Carol
       4 |          20 |    Dave  |              Gina |       Elsa, Fred
       8 |          30 |   Helen  |     Ida, Joe, Ken |             NULL

到目前为止，我的努力：

SELECT 
    sub.person_id, sub.membership_id, sub.first_name, 
    STRING_AGG (sub.dependant, ',')
FROM
    (SELECT 
         person.person_id, person_membership.membership_id, 
         person.first_name, p.first_name AS 'dependant', 
         DATEDIFF(yy, CONVERT(DATETIME, p.dob), GETDATE()) AS 'age'
     FROM   
         person
     LEFT JOIN 
         person_membership ON person.person_id = person_membership.person_id
     LEFT JOIN 
         memship  ON person_membership.membership_id = memship.membership_id
     LEFT JOIN 
         person_membership pm ON person_membership.membership_id = pm.membership_id AND pm.relationship > 1
     LEFT JOIN 
         person p ON pm.person_id = p.person_id
     WHERE 
         person_membership.relationship = 1) as sub
GROUP BY 
    sub.person_id, sub.membership_id, sub.first_name

我不知道如何将CASE WHEN与STRING_AGG一起使用。

当我尝试类似的东西

"CASE WHEN age < 10 THEN STRING_AGG (sub.dependant, ',') ELSE NULL END as 'Under 10'"

服务器正确地抗议

包含在聚合函数或GROUP BY子句中

但是按照它分组当然也不能解决问题，所以我缺少一个窍门。另外，我敢肯定，可以以更简单的方式编写主查询本身。

编辑-解决方案

正如@Gserg正确指出的那样，并且我在发布问题后已经意识到，解决方案非常简单，它要求在STRING_AGG中使用CASE WHEN，而不是相反。 h。

string_agg(case when age < 10 then sub.dependant else null end, ', ') as 'Under 10'

仍在寻找建议和想法，以改进我的原始查询。

Answer 1

使用iif函数针对单个条件最大化。

SELECT sub.person_id, sub.membership_id, sub.first_name, 
    STRING_AGG (iif(age < 10, sub.dependant, null), ',') 'Under 10'
FROM   (SELECT person.person_id, person_membership.membership_id, person.first_name, p.first_name AS 'dependant', 
        DATEDIFF(yy,CONVERT(DATETIME, p.dob),GETDATE()) AS 'age'

        FROM   person

        LEFT JOIN person_membership ON person.person_id = person_membership.person_id
        LEFT JOIN person_membership  memship  ON person_membership.membership_id = memship.membership_id
        LEFT JOIN person_membership pm ON person_membership.membership_id = pm.membership_id AND pm.relationship > 1
        LEFT JOIN person p             ON pm.person_id = p.person_id
        WHERE person_membership.relationship = 1) as sub
GROUP BY sub.person_id, sub.membership_id, sub.first_name

STRING_AGG，CASE WHEN

1 个答案: