如何在mysql表中选择所有大于日期的记录？

Question

My table: ( validfrom and validto and 03JUL2014 below all are of type DATE , not STRING )

id   type    validfrom    validto 
-------------------------------------
1   "APPLE"   10MAY2011    15APR2012
1   "APPLE"   19APR2012    15APR2017
1   "ORANGE"   10MAY2011    15APR2012
1   "ORANGE"   19APR2012    15AUG2018
1   "PEAR"   10MAY2012    15APR2013
1   "PEAR"   10MAY2011    15APR2018
2   "APPLE"   10MAY2011    15APR2017
2   "APPLE"   10MAY2011    15APR2012
2   "ORANGE"   10MAY2011    15APR2012
2   "ORANGE"   10MAY2011    15APR2015
2   "ORANGE"   10MAY2007    15APR2019
2   "PEAR"   10MAY2003    15APR2012
2   "PEAR"   10MAY2006    15APR2022

I want to retrieve all records where id is 2 and validto is greater than 03JUL2014 .

Thus I must get:

2   "APPLE"   10MAY2011    15APR2017
2   "ORANGE"   10MAY2007    15APR2019
2   "PEAR"   10MAY2006    15APR2022

I tried:

SELECT type, validfrom, MAX(validto) FROM crew_qualifications 
               WHERE id=2 AND validto >= 03JUL2014

It only gave me:

2   "PEAR"   10MAY2006    15APR2022

Answer 1

In your query you write MAX(validto) then it return only those row which has validto value is max

you remove MAX(validto) then it return the proper ans

Answer 2

First find the maximum date per type in the given time range. Then join with your table again to get the complete record.

select cq.type, cq.validfrom, cq.validto
from crew_qualifications cq
join
(
  select type, max(validto) as validto
  from crew_qualifications 
  where validto >= '2014-07-03'
  group by type
) wanted_cq
on wanted_cq.type = cq.type and wanted_cq.validto = cq.validto;

As to your query: You select MAX(validto) without groups, so you get the MAX(validto) for the whole table, ie one record. Along with the MAX(validto) you get a random type and a random validfrom from your table, as you don't specify which type and validfrom you want to see (the minimum, maximum, or the average). (Besides validto >= 03JUL2014 should give you a syntax error. I'm surprised it obviously doesn't. I don't know, how MySQL interprets this here.)