合并两个表中不同列的行?

Combine rows from two tables with different columns?

我很难理解这个问题。我相信它正在发生,因为我正在加入基于同一列 (user_id) 的两个单独的表,但我不知道如何修复它,因为两个表之间唯一的共同点是 user_id列。

这里是查询。

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.datedata_new 列连接到或 "appended" 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中?我认为 UNION 可以解决这个问题,但我不能那样做,因为表几乎相同但不共享 message 列。

正如问题所怀疑的那样,我需要的是 UNION。诀窍是在 users_data_new 中创建一个空列以匹配 users_data_existing。我在排序时也遇到了挑战,所以我也会把它包括在这里。

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