Question

我很难绕过这一个。我相信它正在发生，因为我正在基于相同的列（user_id）加入两个单独的表，但我不知道如何解决它，因为两个表之间唯一的共同点是user_id列。

这是查询。

SELECT users_data_existing.`date`,`message`,`action`,`status`,`data`,
users_data_new.`date`,`data_new`
FROM users_data_existing
INNER JOIN users_data_action USING (action_id)
INNER JOIN users_data_status_user USING (status_user_id)
INNER JOIN `users` USING (user_id)
INNER JOIN users_data_new USING (user_id)
INNER JOIN data ON users_data_existing.`data_id` = data.`id`
WHERE users_data_existing.`user_id` = 2
ORDER BY users_data_existing.`date`,users_data_new.`date` DESC

结果是，users_data_new.date和data_new列被连接或“追加”到前面的行。

+----------+-----------+-----------+-----------+-----------+----------+-----------+
|   date   |  message  |   action  |   status  |    data   |   date   |  data_new |
+----------+-----------+-----------+-----------+-----------+----------+-----------+
|2011-01-01|    data   |    data   |    data   |    data   |2011-01-02| data_new  |
-----------------------------------------------------------------------------------
|2011-01-01|    data   |    data   |    data   |    data   |2011-01-03| data_new1 |
-----------------------------------------------------------------------------------

               REPEATS PATTERN FOR TOTAL RECORDS IN users_data_new TABLE 

+----------+-----------+-----------+-----------+-----------+----------+-----------+
|   date   |  message  |   action  |   status  |    data   |   date   |  data_new |
+----------+-----------+-----------+-----------+-----------+----------+-----------+
|2011-01-01|   data1   |   data1   |   data1   |   data1   |2011-01-02| data_new  |
-----------------------------------------------------------------------------------
|2011-01-01|   data1   |   data1   |   data1   |   data1   |2011-01-03| data_new1 |
-----------------------------------------------------------------------------------

但这不是我需要的。如何将最后两列放入单独的行？我认为UNION会解决这个问题，但我不能这样做，因为这些表几乎完全相同但不共享message列。

Answer 1

正如问题中所怀疑的那样，我需要UNION。诀窍是在users_data_new中创建一个空列以匹配users_data_existing。我也有一个排序挑战，所以我也会在这里包括。

(SELECT data_existing.date AS submitdate,status_user.status,action.action,
data.data,data_existing.message
FROM users_data_existing AS data_existing
INNER JOIN users_requested_status_user status_user 
ON data_existing.status_user_id = status_user.status_user_id
INNER JOIN users_requested_action action 
ON data_existing.action_id = action.action_id
INNER JOIN websites data 
ON data_existing.data_id = data.id
ORDER BY data_existing.date DESC) //sorts sub-query

UNION ALL

(SELECT data_new.date AS submitdate,status_user.status,  
action.action,data_new.data_new,'' message //needed to add this last empty column
FROM users_data_new AS data_new
INNER JOIN users_requested_status_user status_user 
ON data_new.status_user_id = status_user.status_user_id
INNER JOIN users_requested_action action 
ON data_new.action_id = action.action_id
ORDER BY data_new.date DESC) //sorts sub-query
ORDER BY submitdate DESC"; //sorts the entire result

请记住，使用日期的别名，关联数组键将是您使用的任何别名。即$result['submitdate']

将两个表中的行与不同的列组合在一起？

1 个答案: