MySQL的。 选择没有1天

Question

Base: name, rate, dt. Every day cost is inserted, I need to know the maximum cost without the first day. The first day of all products is different.

With all dates

SELECT `name`, MAX(rate) AS max
FROM `base`
GROUP BY `name`

This is similar to what I want, but not working.

SELECT `name`, MAX(rate) AS max, MIN(dt) AS min_dt
FROM `base`
WHERE `dt` > `min_dt`
GROUP BY `name`

Eample base

skirt, 6, 2018-10-10 00:00:00 
skirt, 7, 2018-10-11 00:00:00 
cap, 7, 2018-10-11 00:00:00 
skirt, 8, 2018-10-12 00:00:00 
cap, 6, 2018-10-12 00:00:00 

Need

skirt, 8
cap, 6

Answer 1

One approach is to use an inline view to get the min_dt for each name. Then we can join and exclude the minimum date rows

Something like this:

SELECT b.name
     , MAX(b.rate) AS `max`
  FROM ( SELECT d.name 
              , MIN(d.dt) AS min_dt
           FROM `base` d
          GROUP
             BY d.name
       ) m
  JOIN `base` b
    ON b.dt   > m.min_dt
   AND b.name = m.name 
 GROUP
    BY b.name

There are other query patterns that will achieve an equivalent result. My preference would to avoid a correlated subquery, but something like this would also return the specified result:

SELECT b.name
     , MAX(b.rate) AS `max`
  FROM `base` b
 WHERE b.dt > ( SELECT MIN(d.dt) 
                  FROM `base` d
                 WHERE d.name = b.name 
              )
 GROUP
    BY b.name

(With both of these query forms, if there is only row in base for a given name , the query will not return a row for that name .)

Answer 2

You should try to exclude the min dt using subqueries:

SELECT name, MAX(rate) AS max
FROM (SELECT name, rate, dt
      FROM   base B
      WHERE dt NOT IN (SELECT MIN(dt) FROM base WHERE name=B.name ))  as A 
GROUP BY name

Fiddle here