Asterisk-Java以编程方式定义Asterisk版本
[英]Asterisk-Java define Asterisk version programmatically
问题描述
I'm using Asterisk Java v 2.0.4. 
我正在使用Asterisk Java v 2.0.4。
I'm using ManagerConnection to connect to a Grandstream UCM6202 that is based on a Asterisk 13 version. 我正在使用ManagerConnection连接到基于Asterisk 13版本的Grandstream UCM6202 。
When I run my application I get this log: 当我运行我的应用程序时，我得到以下日志：
main] o.a.m.internal.ManagerConnectionImpl : Successfully logged in
2019-05-03 15:27:44.986 ERROR 17108 --- [ main] o.a.m.internal.ManagerConnectionImpl : Unable to determine asterisk version, assuming 1.6... you should expect problems to follow.
2019-05-03 15:27:44.987 INFO 17108 --- [ main] o.a.m.internal.ManagerConnectionImpl : Determined Asterisk version: Asterisk 1.6
is there a way to define programmatically Asterisk version? 有没有办法以编程方式定义星号版本？
1 个解决方案
解决方案1
1 2019-05-03 19:37:54
No, Asterisk Java does not have a way for you to tell it which version of Asterisk it is connecting to. 不可以，Asterisk Java没有办法告诉您它连接到哪个版本的Asterisk。 Asterisk Java determines the version of Asterisk by interacting directly with Asterisk. Asterisk Java通过直接与Asterisk交互来确定Asterisk的版本。
Asterisk Java is open source though so you always have the option to suggest changes to the maintainers/developers or fork the project and customize it for your own use. 尽管Asterisk Java是开源的，所以您始终可以选择向维护者/开发人员提出更改建议，或者派生该项目并对其进行自定义以供您自己使用。 It looks like the version detection code is all centralized in one place: https://github.com/asterisk-java/asterisk-java/blob/master/src/main/java/org/asteriskjava/AsteriskVersion.java#L200 看起来版本检测代码全部集中在一个地方： https : //github.com/asterisk-java/asterisk-java/blob/master/src/main/java/org/asteriskjava/AsteriskVersion.java#L200
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