Mysql中的多个where子句

Question

使用以下查询时，我得到的结果不正确。在处理子句和HAVING中的多个where时，如何解决这个问题。

SELECT * 
FROM  `otc_employee_qualifications` 
WHERE  `emp_qualifctn_type` 
IN (
'26',  '27'
)
AND  `qualification_value` 
IN (
'62', '64','65'
)
AND  `qualification_mark` >=  '10'
GROUP BY  `employee_id` 
HAVING COUNT( DISTINCT  `emp_qualifctn_type` ) =  '2'
AND  COUNT( DISTINCT  `qualification_value` ) =  '3'
LIMIT 0 , 30

PHPMYADMIN

在这里，我需要获得满足emp_qualifctn_type（27和26）以及（＆＃39; 62＆＃39;，＆＃39; 64＆＃39;，＆＃39; 65＆}的所有资格值的员工ID ＃39）

Answer 1

我在查询中发现了错误。我应该在查询中使用'OR'来组合HAVING子句中的多个案例。

SELECT *
FROM  `otc_employee_qualifications` 
WHERE  `emp_qualifctn_type` IN ('26',  '27')
AND  `qualification_value` IN ('62', '64','65')
AND  `qualification_mark` >=  '10'
GROUP BY  `employee_id` 
HAVING COUNT( DISTINCT  `emp_qualifctn_type` ) =  '2'
OR  COUNT( DISTINCT  `qualification_value` ) =  '3'

这里的结果是Null，这是真的。

Answer 2

更新评论I want to get all employee_ids say, who have both emp_qualifctn_type (27 , 29) and both qualification_value (62,50). Here the result would be 2.。检查这个=＆gt;

SELECT * 
FROM  `otc_employee_qualifications` 
WHERE  `emp_qualifctn_type` 
IN (27,  29)
AND  `qualification_value` 
IN (62, 50)
AND  `qualification_mark` >=  10
GROUP BY  `employee_id`
HAVING count(DISTINCT emp_qualifctn_type) >= 2 
LIMIT 0 , 30

Answer 3

根据上述数据，没有员工满足您的评论的所有值我想让所有employee_ids说明，谁同时拥有emp_qualifctn_type（27,29）和qualified_value（62,50）

例如对于2，第一行在emp_qualifctn_type和qualification_value上都匹配，但第二行仅在emp_qualifctn_type上匹配，而第三行仅在qualified_value上匹配。所以2对所有人都没有匹配。

为了获得你想要的东西，我认为你需要匹配一个或另一个而不是AND，然后你可以检查计数： -

SELECT employee_id 
FROM  `otc_employee_qualifications` 
WHERE  (`emp_qualifctn_type` IN (27,  29)
OR  `qualification_value` IN (50, 62))
AND  `qualification_mark` >=  '10'
GROUP BY  `employee_id` 
HAVING COUNT( DISTINCT  `emp_qualifctn_type` ) =  2
AND  COUNT( DISTINCT  `qualification_value` ) =  2