Php 如何使用和条件选择几个值？

我有一个例子 当我做的时候，我需要空的结果

SELECT * FROM prods 
WHERE prod_id = 3 and (fet_id = 1 and fet_id = 5)

但该代码不起作用（fet_id=1和fet_id=5）或

（fet_id=1和fet_id=3）

“Where in”是最好的方法，但我需要“Where in”（1、2和3）

默认情况下，
“其中”
作为
（1或2或3）

当我选择fet_id=1和fet_id=3时，我需要两个结果
当我选择fet_id=1和fet_id=5时，我不需要任何结果

SELECT prod_id FROM prods GROUP by prod_id
HAVING (SUM(fet_id=1)>0
AND SUM(fet_id=5)>0)
OR 
(SUM(fet_id=1)>0
AND SUM(fet_id=3)>0)

在筛选行的位置具有筛选器组。
您可能需要

SELECT prod_id FROM prods GROUP by prod_id
HAVING (SUM(fet_id=1)>0
AND SUM(fet_id=5)>0)
OR 
(SUM(fet_id=1)>0
AND SUM(fet_id=3)>0)

将筛选器组设置为筛选器行。
尝试以下操作：

SELECT p1.id AS id1, p2.id AS id2
FROM `prods` p1 LEFT JOIN `prods` p2 ON p1.`prod_id` =  p2.`prod_id`
WHERE p1.`fet_id` = '1' AND p2.`fet_id` = '3' 

输出

  id1   id2
  1      3

输入fet_id=1和fet_id=5

SELECT p1.id AS id1, p2.id AS id2
FROM `prods` p1 LEFT JOIN `prods` p2 ON p1.`prod_id` =  p2.`prod_id`
WHERE p1.`fet_id` = '1' AND p2.`fet_id` = '5' 

输出：无结果
尝试以下操作：

SELECT p1.id AS id1, p2.id AS id2
FROM `prods` p1 LEFT JOIN `prods` p2 ON p1.`prod_id` =  p2.`prod_id`
WHERE p1.`fet_id` = '1' AND p2.`fet_id` = '3' 

输出

  id1   id2
  1      3

输入fet_id=1和fet_id=5

SELECT p1.id AS id1, p2.id AS id2
FROM `prods` p1 LEFT JOIN `prods` p2 ON p1.`prod_id` =  p2.`prod_id`
WHERE p1.`fet_id` = '1' AND p2.`fet_id` = '5' 

输出：无结果
，如果需要两个以上的值，则另一个结果

SELECT res.* FROM (SELECT 3 as total,
(SELECT COUNT(*) FROM `prods` as t1 WHERE t1.`fet_id` = 1 OR  t1.`fet_id` = 2 OR t1.`fet_id` = 5) as current,
t2.*
FROM `prods` as t2 WHERE t2.`fet_id` in (1,2,5)) as res WHERE res.total = res.current

如果需要两个以上的值，则另一个

SELECT res.* FROM (SELECT 3 as total,
(SELECT COUNT(*) FROM `prods` as t1 WHERE t1.`fet_id` = 1 OR  t1.`fet_id` = 2 OR t1.`fet_id` = 5) as current,
t2.*
FROM `prods` as t2 WHERE t2.`fet_id` in (1,2,5)) as res WHERE res.total = res.current

fet_id=1和fet_id=5）无法工作，因为这个条件不可能为真。fet_id=1和fet_id=5）无法工作，因为这个条件不可能为真。是的，就是这样。谢谢你是的，就是这个。非常感谢。