我有3张桌子，这是我的1张桌子，成员：
+---------+ | user_id | +---------+ | 1 | +---------+
and my table 2,Members7:
+---------------+---------------------------+---------+ | TotalWorkYear | JobScope | user_id | +---------------+---------------------------+---------+ | 1 | Financial;pharmacy | 1 | | 6 | Doctor/Diagnosis;pharmacy | 1 | | 10 | Accounting;pharmacy | 1 | +---------------+---------------------------+---------+
And table 3,Members7_2:
+-----------------+----------------+---------+ | TotalWorkYear_2 | KnowledgeSkill | user_id | +-----------------+----------------+---------+ | 1 | abc;def; | 1 | | 6 | vw;xyz;def | 1 | | 10 | vw;xyz; | 1 | +-----------------+----------------+---------+
I need to find out the user_id that regexp 'pharmacy' in JobScope of Members7,and the total years of JobScope that regexp pharmacy is >=0,and also regexp 'def' in KnowledgeSkill of Members7_2,and the total years of KnowledgeSkill that regexp 'def' is >=0,
so now we can see that:
TotalWorkYear = 17
TotalWorkYear_2 = 7
user_id = '1'
这是我的查询：
SELECT Members.user_id，
SUM（IF（Members7.JobScope
REGEXP'Pharmacy'，Members7.TotalWorkYear，0））年JobScope，Members7.JobScope AS JobScope，
SUM（IF（Member7_2.KnowledgeSkill
REGEXP'def'，Member7_2.TotalWorkYear_2，0））yearsSkills，Members7_2.KnowledgeSkill AS KnowledgeSkill
来自会员
左加入Members7 ON Members.user_id = Members7.user_id
左加入Member7_2 ON Members.user_id = Members7_2.user_id
GROUP BY JobScope REGEXP'药房'，
KnowledgeSkill REGEXP'def'，user_id
拥有
JobScope REGEXP“药房”
AND yearsJobScope> = 0
AND KnowledgeSkill REGEXP'def'
AND yearsSkill> = 0
但输出为：
+ --------------- ++ ------------ +
| yearsJobScope |年技能|
+ --------------- ++ ------------ +
| 17 | 18 |
+ --------------- ++ ------------ +
如果我在HAVING下删除了REGEXP，我会看到：
+ --------- + --------------- + ----------------------- ---------------------- + ------------ + -------------- -+
| user_id | yearsJobScope |工作范围|年技能|知识技能|
+ --------- + --------------- + ----------------------- ---------------------- + ------------ + -------------- -+
| 1 | 1 |审计与税务;银行/金融;制药6 |大众; xyz; def |
| 1 | 6 |医生/诊断;药学| 6 |大众; xyz; def |
| 1 | 10 |通用/成本会计；药学| 6 |大众; xyz; def |
+ --------- + --------------- + ----------------------- ---------------------- + ------------ + -------------- -+
似乎只是对3行的
TotalWorkYears求和。有什么办法解决吗？
最佳答案
您与多个表的左连接将使一些技能计数不止一次。重写查询的简单方法是使用其他表的子查询。
SELECT m.user_id, yearsjobscope, yearsskill
FROM members m
JOIN (
SELECT user_id, SUM(totalworkyear) yearsjobscope
FROM members7 WHERE jobscope REGEXP 'Pharmacy'
GROUP BY user_id) m7
ON m.user_id = m7.user_id
JOIN (
SELECT user_id, SUM(totalworkyear_2) yearsskill
FROM members7_2 WHERE KnowledgeSkill REGEXP 'def'
GROUP BY user_id) m7_2
ON m.user_id = m7_2.user_id
An SQLfiddle to test with。
最后，如果表变大，
REGEX并不是一个好方法，那么您至少要将用分号分隔的值分成几行，以便索引/精确匹配可以覆盖它们。关于mysql - REGEX特定字符的LEFT JOIN表和SUM值，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/17563693/