Question

示例记录：

|name  |price |source|lastest_update|
|name A| 20.00|att   |04/10/2019 00:00:00|
|name A| 30.00|att   |04/11/2019 02:00:00|
|name A| 50.00|sprint|04/10/2019 01:00:00|
|name A| 40.00|sprint|04/11/2019 21:00:00|

基本上，如果我们使用“按名称分组”分组，价格将是记录中的第一个，则是20美元，但我想根据lastest_update（日期）获得最高价格。结果将是：

|name  |att_price|sprint_price|
|name A|  30.00  |  40.00     |

我的查询

SELECT 
MAX(WHEN source = 'att' THEN price ELSE 0 END) as att_price,
MAX(WHEN source = 'sprint' THEN price ELSE 0 END) as sprint_price
FROM table GROUP BY name;

非常感谢您。

Answer 1

我想您可以对LIMIT使用简单的相关查询：

SELECT name
     , (SELECT price FROM t AS x WHERE name = t.name AND source = 'att'    ORDER BY lastest_update DESC LIMIT 1) AS att_price
     , (SELECT price FROM t AS x WHERE name = t.name AND source = 'sprint' ORDER BY lastest_update DESC LIMIT 1) AS sprint_price
FROM t
GROUP BY name

或者可能是双GROUP BY：

SELECT t.name
     , MAX(CASE WHEN t.lastest_update = a.att_date    THEN price END) AS att_price
     , MAX(CASE WHEN t.lastest_update = a.sprint_date THEN price END) AS sprint_price
FROM t
JOIN (
    SELECT name
         , MAX(CASE WHEN source = 'att'    THEN lastest_update END) AS att_date
         , MAX(CASE WHEN source = 'sprint' THEN lastest_update END) AS sprint_date
    FROM t
    GROUP BY name
) AS a ON t.name = a.name
GROUP BY t.name

Answer 2

从时间戳列date中提取latest_update，然后应用max()获取最新日期。现在，您只需添加一个where条件即可过滤仅具有此最大日期的行。

select name,
MAX(case WHEN source = 'att' THEN price ELSE 0 END) as att_price,
MAX(case WHEN source = 'sprint' THEN price ELSE 0 END) as sprint_price
FROM test 
where date(latest_update) = (select max(date(latest_update)) from test)
GROUP BY name;

演示： https://www.db-fiddle.com/f/43Uy7ocCKQRqJSaYHGcyRq/0

更新：

由于您需要根据单独的来源source列为每个latest_update进行分组，因此可以使用以下SQL：

select t1.name,
    max(
        case 
            when t1.source = 'att' then t1.price
            else 0
        end
      ) as att_price,
     max(
        case 
            when t1.source = 'sprint' then t1.price
            else 0
        end
      ) as sprint_price
from test t1
      inner join (
        select name,source,max(latest_update) as latest_update
        from test
        group by name,source) t2
      on t1.name = t2.name and t1.source = t2.source
      and t1.latest_update  = t2.latest_update
group by t1.name;

演示： https://www.db-fiddle.com/f/qwBxizWooVG7AMwqKjewP/0

如何根据查询中的最新日期获取最高价格

2 个答案: