MYSQL(在某些条件下为SELECT)
内容导读
互联网集市收集整理的这篇技术教程文章主要介绍了MYSQL(在某些条件下为SELECT),小编现在分享给大家,供广大互联网技能从业者学习和参考。文章包含2080字,纯文字阅读大概需要3分钟。
内容图文
我是MYSQL初学者,但查询有问题:
我有两个表(calls,callsfailed).
>调用具有这些列(称为“ _number”,“ duration”).
> callfailed有这些列(称为_number,release_reason)
可以说通话表是:
+---------------+----------+
| called_number | duration |
+---------------+----------+
| 1010101 | 13 |
| 1010101 | 18 |
| 1010101 | 20 |
| 2020202 | 50 |
| 2020202 | 20 |
| 3030303 | 10 |
| 4040404 | 30 |
+---------------+----------+
失败呼叫表为:
+---------------+----------------+
| called_number | release_reason |
+---------------+----------------+
| 1010101 | -1 |
| 1010101 | -1 |
| 2020202 | -1 |
| 3030303 | 406 |
| 4040404 | 503 |
| 5050505 | -1 |
| 5050505 | -1 |
| 6060606 | -1 |
+---------------+----------------+
我想选择被叫号码> 1,当持续时间小于25时,或使release_reason = -1.超过1次.
BUT If the caller_number made a duration more than 25, don’t select it even when it has a release_reason = -1.
因此结果将是:
+---------------+-------+
| called_number | count |
+---------------+-------+
| 1010101 | 3 |
| 5050505 | 2 |
+---------------+-------+
我的代码是:
( SELECT called_number, duration, COUNT(*) count
FROM calls
GROUP BY called_number
Having COUNT(called_number) > 1 and duration < 25
)
UNION
( SELECT called_number, release_reason, COUNT(*) count
FROM callsfailed
GROUP BY called_number
Having COUNT(called_number) > 1 and release_reason = -1
)
解决方法:
您可以使用UNION作为派生表,并在上面进行选择
SELECT t1.called_number, t1.count_calls FROM
(
SELECT called_number, MAX(duration) max_duration, COUNT(called_number) count_calls
FROM calls
GROUP BY called_number
HAVING COUNT(called_number) > 1 AND max_duration < 25
UNION
SELECT called_number, release_reason, COUNT(called_number) count_callsfailed
FROM callsfailed
GROUP BY called_number
HAVING COUNT(called_number) > 1 AND release_reason = -1
) t1
GROUP BY t1.called_number
你有你的结果
内容总结
以上是互联网集市为您收集整理的MYSQL(在某些条件下为SELECT)全部内容,希望文章能够帮你解决MYSQL(在某些条件下为SELECT)所遇到的程序开发问题。 如果觉得互联网集市技术教程内容还不错,欢迎将互联网集市网站推荐给程序员好友。
内容备注
版权声明:本文内容由互联网用户自发贡献,该文观点与技术仅代表作者本人。本站仅提供信息存储空间服务,不拥有所有权,不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容, 请发送邮件至 gblab@vip.qq.com 举报,一经查实,本站将立刻删除。
内容手机端
扫描二维码推送至手机访问。