Select 基于 MYSQL 中每个 ID 的日期的最新版本

Question

我有一个table喜欢：

+------------+-------------------+--------------+------------+
| listing_id | transaction_title | image_thumb  | sale_date  |
+------------+-------------------+--------------+------------+
| 226835186  | Title Version 11  | Img Style 11 | 2016-02-08 |
+------------+-------------------+--------------+------------+
| 226835186  | Title Version 11  | Img Style 12 | 2016-02-16 |
+------------+-------------------+--------------+------------+
| 228703248  | Title Version 21  | Img Style 21 | 2016-02-15 |
+------------+-------------------+--------------+------------+
| 228703248  | Title Version 22  | Img Style 22 | 2016-02-17 |
+------------+-------------------+--------------+------------+
| 228703248  | Title Version 23  | Img Style 21 | 2016-02-16 |
+------------+-------------------+--------------+------------+
| 230105831  | Title Version 31  | Img Style 31 | 2016-02-12 |
+------------+-------------------+--------------+------------+
| 230105831  | Title Version 32  | Img Style 31 | 2016-02-06 |
+------------+-------------------+--------------+------------+

我正在尝试使用最新使用的 transaction_title 和 image_thumb 版本查询不同的 listing_id。对于上面的 table 查询输出将是：

+------------+-------------------+--------------+------------+
| listing_id | transaction_title | image_thumb  | sale_date  |
+------------+-------------------+--------------+------------+
| 226835186  | Title Version 11  | Img Style 12 | 2016-02-16 |
+------------+-------------------+--------------+------------+
| 228703248  | Title Version 22  | Img Style 22 | 2016-02-17 |
+------------+-------------------+--------------+------------+
| 230105831  | Title Version 31  | Img Style 31 | 2016-02-12 |
+------------+-------------------+--------------+------------+

我尝试了 select distinct, num_rows and max() 的不同组合，但无法获得想要的结果。

我最近尝试过：

SELECT
    listing_id,transaction_title,image_thumb,sale_date
FROM (
    SELECT * FROM sales
    ORDER BY sale_date DESC
) AS transaction_title
GROUP BY listing_id

请帮忙！

Answer 1

您可以使用相关查询 select 每个查询的最大日期，如下所示：

SELECT listing_id,transaction_title,image_thumb,sale_date
FROM sales t
WHERE (listing_id,sale_date) in (select s.listing_id,max(s.sale_date)
                                 from sales s
                                 where t.listing_id = s.listing_id
                                 group by s.listing_id)

Answer 2

先用子查询得到listing_id最大的sale_date然后再用它找出其他列

select * 
  from table t0 
 where exists
   (select 1 from 
      (select listing_id,max(sale_date) as sale_date 
         from table group by listing_id) t1 
        where t1.listing_id = t0.listing_id and t1.sale_date = t0.sale_date)

Answer 3

你可以使用相关子查询得到预期的结果。

SELECT
    s1.listing_id, s1.transaction_title, s1.image_thumb, s1.sale_date
FROM sales s1
WHERE ( s1.listing_id, s1.transaction_title, s1.image_thumb, s1.sale_date ) IN   
                (SELECT s2.listing_id, s2.transaction_title, s2.image_thumb, s2.sale_date
                 FROM sales s2
                 WHERE s2.listing_id = s1.listing_id
                 ORDER BY s2.sale_date DESC
                 LIMIT 1);

Answer 4

因为没有定义最大值之间的相关性，您可以简单地获取每一列的最大值：

SELECT
    listing_id,
    MAX(transaction_title) transaction_title,
    MAX(image_thumb) image_thumb,
    MAX(sale_date) sale_date
FROM
    sales
GROUP BY
    listing_id

Answer 5

您可以使用包含每个 listing_id 的最大日期的派生 table。如果你 INNER JOIN 这个 table 你可以得到预期的结果集：

select t1.listing_id, transaction_title, image_thumb, sale_date
from mytable as t1
inner join (
   select listing_id, max(sale_date) max_date
   from mytable
   group by listing_id
) as t2 on t1.listing_id = t2.listing_id and sale_date = max_date

Answer 6

您可以使用 row_number 基本方法，数据首先按 listing_id 和 sale_date 降序排列，然后选择 row_number 1 的行。这将为您提供所需的数据集。这种方法的查询模板如下：

SELECT  INVW.listing_id, INVW.transaction_title, INVW.image_thumb, INVW.sale_date 
FROM (
    SELECT  listing_id, transaction_title, image_thumb, sale_date
        ,@rank := if(@listing_id = listing_id or listing_id is null, @rank + 1, 1) as row_number 
        ,@listing_id := listing_id as dummy
    FROM <###REPLACE_ME_WITH_TABLE_NAME###>, (select @rank := 0,@listing_id := '') rank
    ORDER BY listing_id,sale_date DESC 
) INVW where INVW.row_number = 1;

Select 基于 MYSQL 中每个 ID 的日期的最新版本

Select the latest version based on date for each ID in MYSQL

mysql

sql

max

distinct

mysql-num-rows