MYSQL - 如何选择 CASE(两个表，如果一个不存在，请检查另一个)

Question

我拥有的是 mysql 数据库中的两个表。

一个表包含值为 "example@example.com"

的字段

其他表没有...

我想做的是检查两个表中是否有一个字段的值与 "example@example.com" 匹配。

这是我的查询，它不起作用:

(#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 13)

SELECT CASE WHEN
EXISTS (
    SELECT address, suburb, city, postCode, province
    FROM user_postal
    INNER JOIN user_info ON user_postal.id = user_info.id
    WHERE user_info.emailContact = 'example@example.com'
)
OR EXISTS (
    SELECT address, suburb, city, postCode, province
    FROM user_postal
    INNER JOIN user_business_info ON user_postal.id = user_business_info.id
    WHERE user_business_info.emailContact = 'example@example.com'
)

我想返回字段 address, suburb, city, postCode, province 与电子邮件 example@example.com 用户 ID 匹配的字段值>。 (这可能在两个表中的任何一个中，但肯定在一个或另一个中)

谁能告诉我我哪里出了问题？在这方面的任何信息将不胜感激。谢谢!

Answer 1

这个怎么样:

SELECT address, suburb, city, postCode, province
FROM user_postal
INNER JOIN user_info ON user_postal.id = user_info.id
WHERE user_info.emailContact = 'example@example.com'

UNION

SELECT address, suburb, city, postCode, province
FROM user_postal
INNER JOIN user_business_info ON user_postal.id = user_business_info.id
WHERE user_business_info.emailContact = 'example@example.com'