php - 基于此表，如何获得预期的输出 mysql php？

Question

这是一种挑战。预期输出，所有水果都来自纽约州或 ak 州，如果标记好 = 价格 10 和坏 = 价格 20 和 OK = 价格 40。所以，来自纽约州的苹果，标记好和坏匹配价格 10 和 20，和苹果没有标记OK行，所以不需要比较。 Banana mark bad = price 20, but OK is not equal price 40, 所以我会排除香蕉。 pear match OK = price 40 and state=ny 并且没有好坏标记行。因此，我的输出应该是苹果和梨。我应该怎么做？

<?php
/*
fruit  state  price   mark
apple   ny      10    good
apple   ny      20    bad
banana  ny      20    bad
banana  ny      30    OK
pear    ny      40    OK
berry   pa      10    good
*/

/*this is my best query but fail to do what I expected*/
SELECT fruit FROM table WHERE state IN ('ny','ak')  AND mark IN ('good',  'bad', 'OK') AND price IN ( '10','20','40') GROUP BY fruit	
?>

Answer 1

你应该在哪里和或条件

你应该使用一对 not in

select distinct fruit from  my_table 
where fruit not in (select distinct fruit from my_table 
     where (mark, price) not in (select 'good', 10 from dual 
                            union 
                            select 'bad, 20 from dual
                            union 
                            select 'OK', 40 from dual))
state IN ('ny','ak') ;