Select 在一些 "left joins" 之后用最新的日期时间字段完成行

Question

是的，好像回答了很多次，但我都试过了。

最相似的Whosebug的post肯定是：但是这里的主要区别是我需要在之前执行一些左连接之后进行分组。

情况是这样的:

我有 3 个 table（交易、支持和 transaction_support 链接前 2 个 table）

create TABLE `transaction`
(
  id INT,
  date_time DATE,
  notes TEXT,
  PRIMARY KEY (id)
);

create TABLE `support`
(
  id int,
  support_number int ,
  PRIMARY KEY (id)
);

create TABLE `transaction_support`
(
  id INT,
  transaction_id int,
  support_id int,  
  PRIMARY KEY (id), 
  FOREIGN KEY (transaction_id) REFERENCES transaction(id),  
  FOREIGN KEY (support_id) REFERENCES support(id)
);

INSERT INTO `support` values (1, 1111);
INSERT INTO `support` values (2, 2222);
INSERT INTO `support` values (3, 3333);

INSERT INTO `transaction` values (1, '1996-06-28 00:00:00', 'Old data, we shouln''t see it');
INSERT INTO `transaction` values (2, '1996-07-16 00:00:00', 'Old data, we shouln''t see it');
INSERT INTO `transaction` values (3, '2001-04-10 00:00:00', 'Old data, we shouln''t see it');
INSERT INTO `transaction` values (4, '2001-05-14 00:00:00', 'Lastest data from Danny');
INSERT INTO `transaction` values (5, '2001-05-14 00:00:00', 'Lastest data from John');
INSERT INTO `transaction` values (6, '2001-04-10 00:00:00', 'Old data, we shouln''t see it');

INSERT INTO `transaction_support` values (487131, 1, 2);
INSERT INTO `transaction_support` values (488504, 2, 2);
INSERT INTO `transaction_support` values (751126, 3, 2);
INSERT INTO `transaction_support` values (758188, 4, 2);
INSERT INTO `transaction_support` values (4444, 5, 3);
INSERT INTO `transaction_support` values (4445, 6, 3);

这是一个尝试的请求：

SELECT s.id AS s_id, t.*, MAX(t.date_time) AS `this date is good`
FROM support AS s
LEFT JOIN transaction_support AS ts ON ts.support_id = s.id
LEFT JOIN transaction AS t ON ts.transaction_id = t.id
GROUP BY ts.support_id

再次尝试子查询：

SELECT s.id as support_id, t.*, sub.*
FROM support AS s
LEFT JOIN transaction_support AS ts  ON ts.support_id = s.id 
LEFT JOIN transaction AS t ON ts.transaction_id = t.id 
LEFT JOIN (
    SELECT ts.support_id AS `sub_support_id`,
            t.id AS `sub_transaction_id`,
            MAX(t.date_time) AS `sub_last_date`
    FROM transaction_support AS ts 
    LEFT JOIN transaction AS t ON ts.transaction_id = t.id 
    GROUP BY ts.support_id
 ) sub ON ts.support_id = sub.sub_support_id AND t.date_time = sub.sub_last_date
GROUP BY s.id

预期结果为：

|support_id | transaction_id | transaction_notes       | transaction_date|
|-----------|----------------|-------------------------|-----------------|
| 1         | null           | null                    | null            |
| 2         | 4              | Lastest data from Danny | 2001-05-14      |
| 3         | 5              | Lastest data from John  | 2001-05-14      |

我尝试了很多请求，有和没有子查询，但到目前为止，当我“分组”支持 ID 时，我从未从事务 table 中获得所有最新数据。

但我很确定我需要一个子查询...

这是一个 fiddle : http://sqlfiddle.com/#!9/adc611/20

一些其他类似的 post 我试过了 :

GROUP BY with MAX(DATE)
SQL select only rows with max value on a column
MySQL select MAX(datetime) not returning max value

如果有人能帮我找出解决办法...谢谢！ :)

Answer 1

对于您的 MySql 版本，没有简单的解决方案。
您可以使用 NOT EXISTS 获取每个 support_id 的最新 date_time 的数据，并将 support 加入该结果集：

SELECT s.id AS support_id, 
       x.id AS transaction_id,  
       x.notes AS transaction_notes,
       x.date_time AS transaction_date
FROM support AS s
LEFT JOIN (
  SELECT ts.support_id, t.id, t.notes, t.date_time
  FROM transaction_support ts INNER JOIN transaction t
  ON ts.transaction_id = t.id
  WHERE NOT EXISTS (
    SELECT 1
    FROM transaction_support ts2 INNER JOIN transaction t2
    ON ts2.transaction_id = t2.id
    WHERE ts2.support_id = ts.support_id AND t2.date_time > t.date_time
  )
) AS x ON x.support_id = s.id

参见demo。
结果：

> support_id | transaction_id | transaction_notes       | transaction_date
> ---------: | -------------: | :---------------------- | :---------------
>          1 |           null | null                    | null            
>          2 |              4 | Lastest data from Danny | 2001-05-14      
>          3 |              5 | Lastest data from John  | 2001-05-14

Answer 2

如果您想要每个支持的最新交易，一个选项是在 left join 的 on 子句中使用子查询进行过滤：

select s.*, t.*
from support s
left join (
    select t.*, ts.support_id
    from transaction_support ts 
    inner join transaction t 
        on  t.id = ts.transaction_id
        and t.date_time = (
            select max(t1.date_time)
            from transaction_support ts1
            inner join transaction t1 on t1.id = ts1.transaction_id
            where ts1.support_id = ts.support_id
        )
) t on s.id = t.support_id

Select 在一些 "left joins" 之后用最新的日期时间字段完成行

Select complete row with the latest datetime field after some "left joins"

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