Question

我有两个查询返回以下结果（A）＆amp; （B）说：

SELECT username, ext_num FROM user u 
JOIN extension e 
ON u.id=e.user_id;
+----------+---------+
| username | ext_num |
+----------+---------+
| test     | 2459871 |
+----------+---------+
1 row in set (0.00 sec)

SELECT TIME_TO_SEC(TIMEDIFF(od.created_at, oc.created_at)) as `duration (sec)`, oc.ext_num, oc.destination, oc.created_at, oc.call_id
-> FROM on_connected oc
-> JOIN on_disconnected od ON od.call_id = oc.call_id
-> WHERE oc.ext_num = 2459871\G;
*************************** 1. row ***************************
duration (sec): 4
   ext_num: 2459871
destination: 55544466677
created_at: 2013-08-19 17:11:53
   call_id: 521243ad953e-965inwuz1gku
*************************** 2. row ***************************
duration (sec): 4
   ext_num: 2459871
destination: 55544466677
created_at: 2013-08-20 10:28:48
   call_id: 521336b51225-0w4mkelwpfui
2 rows in set (0.00 sec)

我想加入上面的两个表来返回类似的内容：

+----------------+----------------+-----------------------+---------------------+---------------------------+
| username       | duration (sec) | ext_num | destination | created_at          | call_id                   |
+----------------+----------------+-----------------------+---------------------+---------------------------+
| test           |              4 | 2459871 | 55544466677 | 2013-08-19 17:11:53 | 521243ad953e-965inwuz1gku |
| test           |              4 | 2459871 | 55544466677 | 2013-08-20 10:28:48 | 521336b51225-0w4mkelwpfui |
+----------------+----------------+-----------------------+---------------------+---------------------------+

理论上，如果需要，我可以在“call_id”上返回针对任何特定“ext_num”的所有电话或更精细的报告。

我尝试了什么？我最初想到的是UNION运算符：

(A) UNION (B);

其中（A）在SELECT语句中用NULL值填充，但这会产生不稳定的结果。

+----------------+---------+-------------+---------------------+
| duration (sec) | ext_num | destination | created_at          |
+----------------+---------+-------------+---------------------+
| 4              | 2459871 | 55544466677 | 2013-08-19 17:11:53 |
| 4              | 2459871 | 55544466677 | 2013-08-20 10:28:48 |
| test           | 2459871 | NULL        | NULL                |
+----------------+---------+-------------+---------------------+
3 rows in set (0.01 sec)

Answer 1

试试这个

SELECT * FROM 
(SELECT username, ext_num FROM USER u 
JOIN extension e 
ON u.id=e.user_id) a

INNER JOIN

(SELECT TIME_TO_SEC(TIMEDIFF(od.created_at, oc.created_at)) AS `duration (sec)`, oc.ext_num, oc.destination, oc.created_at, oc.call_id
 FROM on_connected oc
JOIN on_disconnected od ON od.call_id = oc.call_id
WHERE oc.ext_num = 2459871 ) b    
ON (a.ext_num = b.ext_num)

其他方式将所有4个表连接在一起

SELECT u.username TIME_TO_SEC(TIMEDIFF(od.created_at, oc.created_at)) AS `duration (sec)`, oc.ext_num, oc.destination, oc.created_at, oc.call_id
 FROM on_connected oc
JOIN on_disconnected od ON od.call_id = oc.call_id
JOIN extension e ON (oc.ext_num = e.ext_num)
JOIN `user` u ON (u.id=e.user_id)
WHERE oc.ext_num = 2459871

基于参数连接多个选择

1 个答案: