我有四张桌子：acting，writing，film，tech
email | skill
-----------------------
foo@bar.com | acting
foo1@bar.com | acting
foo2@bar.com | acting
... | ...
email | skill
-----------------------
foo@bar.com | writing
foo7@bar.com | writing
foo8@bar.com | writing
... | ...
email | skill
-----------------------
foo4@bar.com | film
foo9@bar.com | film
foo12@bar.com | film
... | ...
email | skill
-----------------------
foo3@bar.com | tech
foo4@bar.com | tech
foo7@bar.com | tech
... | ...
我想创建一个新表skills，以便包含每个电子邮件地址，并将与每个地址相关的技能连接到技能字段中。
email | skill
-----------------------
foo@bar.com | acting, writing
foo1@bar.com | writing, film
foo2@bar.com | acting, film, tech
... | ...
我通过联盟进行了全外连接，但我似乎无法做到正确。
更新：我意识到这些表的设计有多糟糕。我正在使用上表从frankenstein遗留CRM迁移数据，我需要在新表中分隔逗号值，以便将它们导入到新的CRM中。
答案 0 :(得分：3)
SELECT email, GROUP_CONCAT(skill ORDER BY skill) skill
FROM
(
SELECT email, skill FROM acting
UNION
SELECT email, skill FROM writing
UNION
SELECT email, skill FROM film
UNION
SELECT email, skill FROM tech
) x
GROUP BY email