SQL查询两个表并合并结果（略有不同）

Question

我有两个表： users和users_info

users看起来像这样：

+----+----------+-------+
| id | slug     | name  |
+----+----------+-------+
|  1 | theploki | Kris  |
+----+----------+-------+

和users_info看起来像这样：

+----+--------+----------+---------------+
| id | parent | info_key | info_val      |
+----+--------+----------+---------------+
|  1 | 1      | email    | kris@kris.com |
+----+--------+----------+---------------+
|  2 | 1      | age      | 28            |
+----+--------+----------+---------------+

我想SELECT一个user谁拥有user_info电子邮件=“kris@kris.com”
-和-
返回所有具有匹配的users_info.parent = user.id user_info结果
-和-
以可读格式（例如result['email'] = 'kris@kris.com'和result['id'] = 1以及result['name'] = 'Kris'和result['age'] = 28 ）

我确定这已在另一个问题中得到解答，但我一直在搜索和阅读对我而言并不完全有效的答案。

显然，我确保info_key不能与users列名匹配。

我尝试了INNER JOIN，OUTER JOIN，只是简单的JOIN，并且我尝试了完全没有任何JOIN。

这不起作用，但是有点解释了我想做什么：

SELECT * FROM users
WHERE 1=1
AND users.slug='theploki'
AND (SELECT * FROM users_info
    WHERE users_info.parent = users.id
    AND 1=1
    AND users_info.info_key = 'email'
    AND users_info.info_val = 'kris@kris.com'
    GROUP BY users_info.parent)
GROUP BY users.id

有时我不会搜索信息列，有时我不会搜索用户列，这就是为什么我将1=1

也许执行两个查询然后将它们组合起来会更简单？

更新：

好的，这是更新的SQL查询（由scaisEdge提供）：

SELECT users.*, users_info.* FROM users
INNER JOIN users_info on users_info.parent = users.id
where users.id = (SELECT users_info.parent FROM users_info
    WHERE users_info.parent = users.id
    AND users_info.info_val = 'kris@kris.com')

这非常接近，但是返回如下结果集：

+----+----------+-------+----+--------+----------+---------------+
| id | slug     | name  | id | parent | info_key | info_val      |
+----+----------+-------+----+--------+----------+---------------+
|  1 | theploki | Kris  |  1 |  1     | email    | kris@kris.com |
+----+----------+-------+----+--------+----------+---------------+
|  1 | theploki | Kris  |  2 |  1     | age      | 28            |
+----+----------+-------+----+--------+----------+---------------+

但我想要这样的结果集：

+----+----------+-------+---------------+-----+
| id | slug     | name  | email         | age |
+----+----------+-------+---------------+-----+
|  1 | theploki | Kris  | kris@kris.com | 28  |
+----+----------+-------+---------------+-----+

Answer 1

同一父用户应为

SELECT users.*, users_info.* FROM users
INNER JOIN users_info on users_info.parent_id = users.id
where users.id = (SELECT users_info.parent FROM users_info
    WHERE users_info.parent = users.id
    AND users_info.info_val = 'kris@kris.com')

仅作为示例，您可以使用列名和别名更改user。* ord users_info。*

SELECT users.slug,  users.name as name, group_concat(users_info.info_val) FROM users
INNER JOIN users_info on users_info.parent_id = users.id
where users.id = (SELECT users_info.parent FROM users_info
    WHERE users_info.parent = users.id
    AND users_info.info_val = 'kris@kris.com')
 group by users.name

Answer 2

SELECT users_info.info_val AS email 
FROM users JOIN users_info  ON users_info.parent_id = users.id
WHERE users_info.info_key = 'email'

那应该为您想要的结果工作。 提供的另一个答案要求您在编写查询之前知道要查找的结果。 不过，正如我在上面的评论中提到的那样，这是一种结构化数据库的奇怪方法，除非您有特殊原因，否则常规用户信息（例如电子邮件地址）将位于users表中。

但是，如果您打算有一个最小的用户表，然后为他们提供通用的键值编码信息，则此查询将起作用。