Question

我想使用分组（open）的MIN（id）行中exchange, base_currency, quote_currency, DATE(created_at)的值来更新组（last）的MAX（id）的id last open exchange base_curr quote_curr created_at 6 1.11 0.00 ex1 usd yen 2018-07-29 03:00:00 --> update open with 1.14 (value of last from MIN(id) of group) 5 1.09 0.00 ex1 usd yen 2018-07-29 02:00:00 4 1.14 0.00 ex1 usd yen 2018-07-29 01:00:00 3 0.49 0.00 ex2 yen won 2018-07-29 03:00:00 --> update open with 0.50 (value of last from MIN(id) of group) 2 0.51 0.00 ex2 yen won 2018-07-29 02:00:00 1 0.50 0.00 ex2 yen won 2018-07-29 01:00:00值同一组。

last

我了解如何获取组中的所有MIN（id），但不确定如何使用MIN（id）行中的open值来更新{{ 1}}组的MAX（id）值。

MAX（id）或MAX（created_at）将为我提供组的最新行。

SELECT MIN(id) as min_id, last
FROM tickers 
WHERE DATE(created_at) = '2018-07-29' 
GROUP BY exchange, base_currency, quote_currency, DATE(created_at)

Answer 1

您可以在子查询中计算最小值/最大值。然后使用另一个join从最小值行中引入值：

update tickers t join
       (select exchange, base_curr, quote_curr, date(created_at) as created_at_date,
               max(id) as maxid, min(id) as minid
        from tickers t2
        group by exchange, base_curr, quote_curr, date(created_at)
       ) tt
       on tt.maxid = t.id join
       tickers tmin
       on tmin.id = tt.minid
    set t.open = tmin.last;

使用同一组的MIN（id）的`open`值的组的mysql UPDATE MAX值，多行

1 个答案: