MySQL 没有重复数据的最近日期

Question

所以我需要显示我所有的客户和相关的预订号（如果没有预订则为空），而不需要重复的客户。如果客户有很多预订，我只需要显示最近的预订日期。我不明白为什么我的查询不起作用。

这是做了什么：http://sqlfiddle.com/#!9/df0455/19

SELECT c.name, x.number, x.start_date
FROM customer c 
LEFT JOIN 
(SELECT b.customer_id, b.number, b.start_date
 FROM booking b
 INNER JOIN (
    SELECT customer_id, MIN(ABS(TIME_TO_SEC(TIMEDIFF(NOW(), start_date)))) as mindiff
    FROM booking
    GROUP BY customer_id
  ) nearest ON b.customer_id = nearest.customer_id AND ABS(TIME_TO_SEC(TIMEDIFF(NOW(), start_date))) = mindiff
) AS x ON c.id = x.customer_id

其实Paul显示了3次，只需要显示Paul一次，最近的预订号是booking-1 2019-11-05 21:45:00

希望你能帮帮我

Answer 1

您可以使用行限制相关子查询进行过滤：

select c.name, b.number, b.start_date
from customer c
inner join booking b on b.customer_id = c.id
where b.start_date = (
    select b1.start_date
    from booking b1
    where b1.customer_id = b.customer_id
    order by abs(timestampdiff(second, now(), b1.start_date)) 
    limit 1
)

在您的数据库中 Fiddle、this produces:

name    number     start_date
Paul    booking-1   2019-11-05T21:45:00Z
John    booking-3   2019-09-27T21:45:00Z
Morgan  booking-5   2019-09-27T21:45:00Z

如果您还想显示没有预订的客户，那么您可以 left join 并将过滤移动到 join:

的 on 子句

select c.name, b.number, b.start_date
from customer c
left join booking b 
    on b.customer_id = c.id
    and b.start_date = (
        select b1.start_date
        from booking b1
        where b1.customer_id = b.customer_id
        order by abs(timestampdiff(second, now(), b1.start_date)) 
        limit 1
    )

Answer 2

您可以使用NOT EXISTS获取最近的预订并加入客户：

SELECT c.id, c.name, t.number, t.start_date
FROM customer c 
LEFT JOIN (
  SELECT b.* FROM booking b
  WHERE NOT EXISTS (
    SELECT 1 FROM booking
    WHERE customer_id = b.customer_id 
      AND ABS(TIMESTAMPDIFF(SECOND, NOW(), start_date)) < ABS(TIMESTAMPDIFF(SECOND, NOW(), b.start_date))
  )  
) t ON t.customer_id = c.id

参见demo。
结果：

| id  | name   | number    | start_date          |
| --- | ------ | --------- | ------------------- |
| 1   | Paul   | booking-1 | 2019-11-05 21:45:00 |
| 2   | John   | booking-3 | 2019-09-27 21:45:00 |
| 3   | Morgan | booking-5 | 2019-09-27 21:45:00 |
| 4   | Jane   |           |                     |
| 5   | Mike   |           |                     |

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