如何显示 mysql table 中多个条目的数据，这些条目通过单个值匹配与另一个条目连接？
How do you display data from multiple entries in a mysql table that are JOINED with another via a single value match?
我需要 select 并显示来自一对 MySQL table 的信息，但我无法理解语法。具体来说，我需要将 cwd_user table 中的数据与 cwd_user_attribute table 字段中的数据 cwd_user.id == cwd_user_attribute.user_id，但我还需要在一行中显示 cwd_user_attribute table 中多个条目的值。是后者让我望而却步。以下是血淋淋的细节：
给定两个 tables：
mysql (crowd@prod:crowddb)> desc cwd_user;
+---------------------+--------------+------+-----+---------+-------+
| Field | Type | Null | Key | Default | Extra |
+---------------------+--------------+------+-----+---------+-------+
| id | bigint(20) | NO | PRI | NULL | |
| user_name | varchar(255) | NO | | NULL | |
| active | char(1) | NO | MUL | NULL | |
| created_date | datetime | NO | | NULL | |
| updated_date | datetime | NO | | NULL | |
| display_name | varchar(255) | YES | | NULL | |
| directory_id | bigint(20) | NO | MUL | NULL | |
+---------------------+--------------+------+-----+---------+-------+
mysql (crowd@prod:crowddb)> desc cwd_user_attribute;
+-----------------------+--------------+------+-----+---------+-------+
| Field | Type | Null | Key | Default | Extra |
+-----------------------+--------------+------+-----+---------+-------+
| id | bigint(20) | NO | PRI | NULL | |
| user_id | bigint(20) | NO | MUL | NULL | |
| directory_id | bigint(20) | NO | MUL | NULL | |
| attribute_name | varchar(255) | NO | | NULL | |
| attribute_value | varchar(255) | YES | | NULL | |
+-----------------------+--------------+------+-----+---------+-------+
假设 cwd_user_attribute.attribute_name 最多有七个可能的值，我对其中四个感兴趣：lastAuthenticated、Team、Manager Notes。示例：
mysql (crowd@prod:crowddb)> select * from cwd_user_attribute where user_id = (select id from cwd_user where user_name = 'gspinrad');
+---------+---------+--------------+-------------------------+----------------------------------+
| id | user_id | directory_id | attribute_name | attribute_value |
+---------+---------+--------------+-------------------------+----------------------------------+
| 65788 | 32844 | 1 | invalidPasswordAttempts | 0 |
| 65787 | 32844 | 1 | lastAuthenticated | 1473360428804 |
| 65790 | 32844 | 1 | passwordLastChanged | 1374005378040 |
| 65789 | 32844 | 1 | requiresPasswordChange | false |
| 4292909 | 32844 | 1 | Team | Engineering - DevOps |
| 4292910 | 32844 | 1 | Manager | Matt Karaffa |
| 4292911 | 32844 | 1 | Notes | Desk 32:2:11 |
+---------+---------+--------------+-------------------------+----------------------------------+
5 rows in set (0.00 sec)
我可以通过以下查询获得按 lastAuthenticated 排序的用户列表：
SELECT cwd_user.user_name, cwd_user.id, cwd_user.display_name, from_unixtime(cwd_user_attribute.attribute_value/1000) as last_login FROM cwd_user JOIN cwd_directory ON cwd_user.directory_id = cwd_directory.id JOIN cwd_user_attribute ON cwd_user.id = cwd_user_attribute.user_id AND cwd_user_attribute.attribute_name='lastAuthenticated' WHERE DATEDIFF((NOW()), (from_unixtime(cwd_user_attribute.attribute_value/1000))) > 90 and cwd_user.active='T' order by last_login limit 4;
结果：
+-----------------------+---------+-----------------------+---------------------+
| user_name | id | display_name | last_login |
+-----------------------+---------+-----------------------+---------------------+
| jenkins-administrator | 1605636 | Jenkins Administrator | 2011-10-27 17:28:05 |
| sonar-administrator | 1605635 | Sonar Administrator | 2012-02-06 15:59:59 |
| jfelix | 1605690 | Joey Felix | 2012-02-06 19:15:15 |
| kbitters | 3178497 | Kitty Bitters | 2013-09-03 10:09:59 |
我需要在输出中添加的是 cwd_user_attribute.attribute_value 的值，其中 cwd_user_attribute.attribute_name 是团队、经理、and/or 注释。输出看起来像这样：
+-----------------------+---------+-----------------------+-------------------------------------------------------------------+
| user_name | id | display_name | last_login | Team | Manager | Notes |
+-----------------------+---------+-----------------------+-------------------------------------------------------------------+
| jenkins-administrator | 1605636 | Jenkins Administrator | 2011-10-27 17:28:05 | Internal | Internal | |
| sonar-administrator | 1605635 | Sonar Administrator | 2012-02-06 15:59:59 | Internal | Internal | |
| jfelix | 1605690 | Joey Felix | 2012-02-06 19:15:15 | Hardware Eng. | Gary Spinrad | Desk 32:1:51 |
| kbitters | 3178497 | Kitty Bitters | 2013-09-03 10:09:59 | Software QA | Matt Karaffa | Desk 32:2:01 |
+-----------------------+---------+-----------------------+-------------------------------------------------------------------+
您可以通过附加 LEFT JOIN 属性 table 来实现该结果。然后使用 GROUP BY 和聚合 CASE 语句来转换结果（行到列）。
SELECT
cwd_user.user_name,
cwd_user.id,
cwd_user.display_name,
from_unixtime(cwd_user_attribute.attribute_value/1000) as last_login,
MIN(CASE WHEN attr2.attribute_name = 'TEAM' THEN attr2.attribute_value END) as Team,
MIN(CASE WHEN attr2.attribute_name = 'Manager' THEN attr2.attribute_value END) as Manager,
MIN(CASE WHEN attr2.attribute_name = 'Notes' THEN attr2.attribute_value END) as Notes
FROM
cwd_user
JOIN
cwd_user_attribute ON cwd_user.id = cwd_user_attribute.user_id
AND cwd_user_attribute.attribute_name='lastAuthenticated'
LEFT JOIN
cwd_user_attribute attr2 ON cwd_user.id = attr2.user_id
AND attr2.attribute_name IN ('Team', 'Manager', 'Notes')
WHERE
DATEDIFF((NOW()), (from_unixtime(cwd_user_attribute.attribute_value/1000))) > 90
AND cwd_user.active = 'T'
GROUP BY
cwd_user.id
ORDER BY
last_login
LIMIT 4
使用严格模式，您需要在 GROUP BY 子句中列出所有未聚合的列
GROUP BY
cwd_user.user_name,
cwd_user.id,
cwd_user.display_name,
cwd_user_attribute.attribute_value
另一种方法是只使用三个 LEFT JOIN（每个属性名称一个连接）：
SELECT
cwd_user.user_name,
cwd_user.id,
cwd_user.display_name,
from_unixtime(cwd_user_attribute.attribute_value/1000) as last_login,
attr_team.attribute_value as Team,
attr_manager.attribute_value as Manager,
attr_notes.attribute_value as Notes
FROM cwd_user
JOIN cwd_user_attribute
ON cwd_user.id = cwd_user_attribute.user_id
AND cwd_user_attribute.attribute_name='lastAuthenticated'
LEFT JOIN cwd_user_attribute attr_team
ON cwd_user.id = attr2.user_id
AND attr2.attribute_name = 'Team'
LEFT JOIN cwd_user_attribute attr_manager
ON cwd_user.id = attr2.user_id
AND attr2.attribute_name = 'Manager'
LEFT JOIN cwd_user_attribute attr_notes
ON cwd_user.id = attr2.user_id
AND attr2.attribute_name = 'Notes'
WHERE DATEDIFF((NOW()), (from_unixtime(cwd_user_attribute.attribute_value/1000))) > 90
and cwd_user.active='T'
order by last_login limit 4
注意：我已经删除了与目录 table 的连接，因为您似乎不使用它。如果您需要它进行过滤，请再次添加它。
注意 2：您经常用于搜索的一些属性（如 lastAuthenticated）应转换为用户中的索引列 table 以提高搜索性能。
我需要 select 并显示来自一对 MySQL table 的信息，但我无法理解语法。具体来说，我需要将 cwd_user table 中的数据与 cwd_user_attribute table 字段中的数据 cwd_user.id == cwd_user_attribute.user_id，但我还需要在一行中显示 cwd_user_attribute table 中多个条目的值。是后者让我望而却步。以下是血淋淋的细节：
给定两个 tables：
mysql (crowd@prod:crowddb)> desc cwd_user;
+---------------------+--------------+------+-----+---------+-------+
| Field | Type | Null | Key | Default | Extra |
+---------------------+--------------+------+-----+---------+-------+
| id | bigint(20) | NO | PRI | NULL | |
| user_name | varchar(255) | NO | | NULL | |
| active | char(1) | NO | MUL | NULL | |
| created_date | datetime | NO | | NULL | |
| updated_date | datetime | NO | | NULL | |
| display_name | varchar(255) | YES | | NULL | |
| directory_id | bigint(20) | NO | MUL | NULL | |
+---------------------+--------------+------+-----+---------+-------+
mysql (crowd@prod:crowddb)> desc cwd_user_attribute;
+-----------------------+--------------+------+-----+---------+-------+
| Field | Type | Null | Key | Default | Extra |
+-----------------------+--------------+------+-----+---------+-------+
| id | bigint(20) | NO | PRI | NULL | |
| user_id | bigint(20) | NO | MUL | NULL | |
| directory_id | bigint(20) | NO | MUL | NULL | |
| attribute_name | varchar(255) | NO | | NULL | |
| attribute_value | varchar(255) | YES | | NULL | |
+-----------------------+--------------+------+-----+---------+-------+
假设 cwd_user_attribute.attribute_name 最多有七个可能的值，我对其中四个感兴趣：lastAuthenticated、Team、Manager Notes。示例：
mysql (crowd@prod:crowddb)> select * from cwd_user_attribute where user_id = (select id from cwd_user where user_name = 'gspinrad');
+---------+---------+--------------+-------------------------+----------------------------------+
| id | user_id | directory_id | attribute_name | attribute_value |
+---------+---------+--------------+-------------------------+----------------------------------+
| 65788 | 32844 | 1 | invalidPasswordAttempts | 0 |
| 65787 | 32844 | 1 | lastAuthenticated | 1473360428804 |
| 65790 | 32844 | 1 | passwordLastChanged | 1374005378040 |
| 65789 | 32844 | 1 | requiresPasswordChange | false |
| 4292909 | 32844 | 1 | Team | Engineering - DevOps |
| 4292910 | 32844 | 1 | Manager | Matt Karaffa |
| 4292911 | 32844 | 1 | Notes | Desk 32:2:11 |
+---------+---------+--------------+-------------------------+----------------------------------+
5 rows in set (0.00 sec)
我可以通过以下查询获得按 lastAuthenticated 排序的用户列表：
SELECT cwd_user.user_name, cwd_user.id, cwd_user.display_name, from_unixtime(cwd_user_attribute.attribute_value/1000) as last_login FROM cwd_user JOIN cwd_directory ON cwd_user.directory_id = cwd_directory.id JOIN cwd_user_attribute ON cwd_user.id = cwd_user_attribute.user_id AND cwd_user_attribute.attribute_name='lastAuthenticated' WHERE DATEDIFF((NOW()), (from_unixtime(cwd_user_attribute.attribute_value/1000))) > 90 and cwd_user.active='T' order by last_login limit 4;
结果：
+-----------------------+---------+-----------------------+---------------------+
| user_name | id | display_name | last_login |
+-----------------------+---------+-----------------------+---------------------+
| jenkins-administrator | 1605636 | Jenkins Administrator | 2011-10-27 17:28:05 |
| sonar-administrator | 1605635 | Sonar Administrator | 2012-02-06 15:59:59 |
| jfelix | 1605690 | Joey Felix | 2012-02-06 19:15:15 |
| kbitters | 3178497 | Kitty Bitters | 2013-09-03 10:09:59 |
我需要在输出中添加的是 cwd_user_attribute.attribute_value 的值，其中 cwd_user_attribute.attribute_name 是团队、经理、and/or 注释。输出看起来像这样：
+-----------------------+---------+-----------------------+-------------------------------------------------------------------+
| user_name | id | display_name | last_login | Team | Manager | Notes |
+-----------------------+---------+-----------------------+-------------------------------------------------------------------+
| jenkins-administrator | 1605636 | Jenkins Administrator | 2011-10-27 17:28:05 | Internal | Internal | |
| sonar-administrator | 1605635 | Sonar Administrator | 2012-02-06 15:59:59 | Internal | Internal | |
| jfelix | 1605690 | Joey Felix | 2012-02-06 19:15:15 | Hardware Eng. | Gary Spinrad | Desk 32:1:51 |
| kbitters | 3178497 | Kitty Bitters | 2013-09-03 10:09:59 | Software QA | Matt Karaffa | Desk 32:2:01 |
+-----------------------+---------+-----------------------+-------------------------------------------------------------------+
您可以通过附加 LEFT JOIN 属性 table 来实现该结果。然后使用 GROUP BY 和聚合 CASE 语句来转换结果（行到列）。
SELECT
cwd_user.user_name,
cwd_user.id,
cwd_user.display_name,
from_unixtime(cwd_user_attribute.attribute_value/1000) as last_login,
MIN(CASE WHEN attr2.attribute_name = 'TEAM' THEN attr2.attribute_value END) as Team,
MIN(CASE WHEN attr2.attribute_name = 'Manager' THEN attr2.attribute_value END) as Manager,
MIN(CASE WHEN attr2.attribute_name = 'Notes' THEN attr2.attribute_value END) as Notes
FROM
cwd_user
JOIN
cwd_user_attribute ON cwd_user.id = cwd_user_attribute.user_id
AND cwd_user_attribute.attribute_name='lastAuthenticated'
LEFT JOIN
cwd_user_attribute attr2 ON cwd_user.id = attr2.user_id
AND attr2.attribute_name IN ('Team', 'Manager', 'Notes')
WHERE
DATEDIFF((NOW()), (from_unixtime(cwd_user_attribute.attribute_value/1000))) > 90
AND cwd_user.active = 'T'
GROUP BY
cwd_user.id
ORDER BY
last_login
LIMIT 4
使用严格模式，您需要在 GROUP BY 子句中列出所有未聚合的列
GROUP BY
cwd_user.user_name,
cwd_user.id,
cwd_user.display_name,
cwd_user_attribute.attribute_value
另一种方法是只使用三个 LEFT JOIN（每个属性名称一个连接）：
SELECT
cwd_user.user_name,
cwd_user.id,
cwd_user.display_name,
from_unixtime(cwd_user_attribute.attribute_value/1000) as last_login,
attr_team.attribute_value as Team,
attr_manager.attribute_value as Manager,
attr_notes.attribute_value as Notes
FROM cwd_user
JOIN cwd_user_attribute
ON cwd_user.id = cwd_user_attribute.user_id
AND cwd_user_attribute.attribute_name='lastAuthenticated'
LEFT JOIN cwd_user_attribute attr_team
ON cwd_user.id = attr2.user_id
AND attr2.attribute_name = 'Team'
LEFT JOIN cwd_user_attribute attr_manager
ON cwd_user.id = attr2.user_id
AND attr2.attribute_name = 'Manager'
LEFT JOIN cwd_user_attribute attr_notes
ON cwd_user.id = attr2.user_id
AND attr2.attribute_name = 'Notes'
WHERE DATEDIFF((NOW()), (from_unixtime(cwd_user_attribute.attribute_value/1000))) > 90
and cwd_user.active='T'
order by last_login limit 4
注意：我已经删除了与目录 table 的连接，因为您似乎不使用它。如果您需要它进行过滤，请再次添加它。
注意 2：您经常用于搜索的一些属性（如 lastAuthenticated）应转换为用户中的索引列 table 以提高搜索性能。