提问者:小点点

mysql组和订单行


我有下表“persons”,不同行中有相同的人

id |   firstname   |   surname   | date_created
------------------------------------------------------
3  | Nelli         | Schaller    | 2017-08-22 20:57:19
------------------------------------------------------
4  | Carl          | Schaller    | 2019-06-21 08:29:45
------------------------------------------------------
48 | Nelli         | Schaller    | 2020-06-25 13:06:09
------------------------------------------------------   
49 | Carl          | Schaller    | 2020-06-25 13:06:09

我想要得到的是具有最大id/最新的date_created值的所有唯一的Schallers。

我试过了

SELECT id, CONCAT(surname, ", ", firstname) AS person, date_created
FROM persons
WHERE
surname LIKE "schall%"
GROUP by firstname, surname 
ORDER BY date_createdDESC, surname ASC LIMIT 0, 10

但只得到预期的前两个条目(id 3和4),但我需要48和49。 正如在一些评论中提到的,在这种情况下,LIKE语句不是必需的,但在现实生活中,它将是一个自动完成字段的来源,所以我需要LIKE,任何想法,如何管理它?


共3个答案

匿名用户

使用不存在:

SELECT p.id, CONCAT(p.surname, ', ', p.firstname) AS person, p.date_created
FROM persons p
WHERE p.surname LIKE '%schall%'
AND NOT EXISTS (SELECT 1 FROM persons WHERE firstname = p.firstname AND surname = p.surname AND id > p.id)
ORDER BY p.date_created DESC, person

如果选择每个组中最新一个的条件是列date_created,则更改:

...AND id > p.id

...AND date_created > p.date_created

匿名用户

可以将子查询与组一起用于最大id

select t.max_id, t.person, m.date_created
from (
    SELECT max(id) max_id, CONCAT(surname, ", ", firstname) AS person
    FROM persons
    WHERE surname LIKE "schall%"
    ORDER BY date_createdDESC, surname ASC 
    GROUP BY CONCAT(surname, ", ", firstname)

    ) t 
    inner join  persons m ON  CONCAT(m.surname, ", ", m.firstname) = t.person
        and m-id = t.max_id

匿名用户

SELECT p.*
FROM persons p
LEFT JOIN persons p2 ON p2.firstname = p.firstname 
    AND p2.lastname = p.lastname
    AND p2.date_created > p.date_created
WHERE p2.id IS NULL

这是SQL Server的语法,但MySQL可能类似。

我假设您的id字段和date_created字段都不需要检查,因为它是一个标识列,而且对于后者创建的记录来说会更大,但显然要根据您的实际数据进行调整。