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获得LEFT OUTER JOIN的第一行

获得LEFT OUTER JOIN的第一行

尝试

数据源:

CREATE TABLE person
    (person_id int primary key, firstname varchar2(4), lastname varchar2(9))
/
INSERT ALL
    INTO person (person_id, firstname, lastname)
         VALUES (1, 'john', 'lennon')
    INTO person (person_id, firstname, lastname)
         VALUES (2, 'paul', 'mccartney')
SELECT * FROM dual;



CREATE TABLE address
    (person_id int, address_id int primary key, city varchar2(8))
/
INSERT ALL
    INTO address (person_id, address_id, city)
         VALUES (1, 1, 'new york')
    INTO address (person_id, address_id, city)
         VALUES (1, 2, 'england')
    INTO address (person_id, address_id, city)
         VALUES (1, 3, 'japan')
    INTO address (person_id, address_id, city)
         VALUES (2, 4, 'london')
SELECT * FROM dual;

询问:

    select

      p.person_id, p.firstname, p.lastname,

      x.recent_city

    from person p
    left join (

        select person_id,

            min(city) -- can change this to max(city). will work regardless of min/max

            -- important you do this to get the recent: keep(dense_rank last)

            keep(dense_rank last order by address_id) 
               as recent_city

        from address 
        group by person_id


    ) x on x.person_id = p.person_id

实时测试:http//www.sqlfiddle.com/#!4 / 7b1c9 / 2

并非所有数据库都具有与Oracle的KEEP DENSE_RANK窗口化功能类似的功能,您可以改用普通窗口化功能

select

  p.person_id, p.firstname, p.lastname,

  x.recent_city, x.pick_one_only

from person p
left join (

    select

        person_id,

        row_number() over(partition by person_id order by address_id desc) as pick_one_only,
        city as recent_city

    from address



) x on x.person_id = p.person_id and x.pick_one_only = 1

实时测试:http//www.sqlfiddle.com/#!4 / 7b1c9 /48

或使用元组测试,应在不支持窗口功能数据库上工作:

select

  p.person_id, p.firstname, p.lastname,

  x.recent_city

from person p
left join (

    select   
        person_id,city as recent_city    
    from address 
    where (person_id,address_id) in

          (select person_id, max(address_id)
           from address
           group by person_id)



) x on x.person_id = p.person_id

实时测试:http//www.sqlfiddle.com/#!4 / 7b1c9 / 21

并非像上面的代码中那样,所有数据库支持元组测试。您可以改用JOIN:

select

  p.person_id, p.firstname, p.lastname,

  x.recent_city

from person p
left join (

    select

        address.person_id,address.city as recent_city

    from address 
    join 
    (
          select person_id, max(address_id) as recent_id
           from address
           group by person_id
    ) r 
    ON address.person_id = r.person_id
    AND address.address_id = r.recent_id



) x on x.person_id = p.person_id

实时测试:http//www.sqlfiddle.com/#!4 / 7b1c9 / 24

其他 2022/1/1 18:53:25 有449人围观

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