MySQL Query INNER JOIN с псевдоними

В този случай можете да го направите по следния начин;) Просто обикновена опорна таблица.

select
    users.id,
    users.slug,
    users.name,
    max(if(users_info.info_key = 'email', users_info.info_val, null)) as email,
    max(if(users_info.info_key = 'age', users_info.info_val, null)) as age
from users
inner join users_info
on users.id = users_info.parent
group by users.id

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Ако имате динамичен info_key , ще ви трябва динамичен sql, за да направите това, тук ви давам пример.

SET @sql = NULL;
SELECT
  GROUP_CONCAT(DISTINCT
    CONCAT(
      'max(if(users_info.info_key = ''',
      users_info.info_key,
      ''', users_info.info_val, null)) as ',
      users_info.info_key
    )
  ) INTO @sql
FROM users
inner join users_info
on users.id = users_info.parent
;

SET @sql = CONCAT('select users.id, users.slug, users.name, ', @sql, ' FROM users
inner join users_info group by users.id having email = \'[email protected]\'');

PREPARE stmt FROM @sql;
EXECUTE stmt;
DEALLOCATE PREPARE stmt;

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