sip

Can I turn off SRTP when use webrtc

由 你说的曾经没有我的故事 提交于 2020-07-04 16:15:28
问题 Now I test webrtc communicate with SIP Client(sx20) I send invite message with webrtc sdp. but sip client answer has not finger print, and sip client answer is not SRTP just RTP. So I need to turn off SRTP in WEBRTC. Can I do it? and one more question... I think, finger print is encrypt key, so, sender and receiver need to each key? we need two key? sender's key, receive's key? or we need just sender' key? 回答1: You cannot turn off SRTP. It is part of the standard and it will probably never be

Can I turn off SRTP when use webrtc

由 依然范特西╮ 提交于 2020-07-04 16:13:48
问题 Now I test webrtc communicate with SIP Client(sx20) I send invite message with webrtc sdp. but sip client answer has not finger print, and sip client answer is not SRTP just RTP. So I need to turn off SRTP in WEBRTC. Can I do it? and one more question... I think, finger print is encrypt key, so, sender and receiver need to each key? we need two key? sender's key, receive's key? or we need just sender' key? 回答1: You cannot turn off SRTP. It is part of the standard and it will probably never be

Can I turn off SRTP when use webrtc

由 放肆的年华 提交于 2020-07-04 16:12:01
问题 Now I test webrtc communicate with SIP Client(sx20) I send invite message with webrtc sdp. but sip client answer has not finger print, and sip client answer is not SRTP just RTP. So I need to turn off SRTP in WEBRTC. Can I do it? and one more question... I think, finger print is encrypt key, so, sender and receiver need to each key? we need two key? sender's key, receive's key? or we need just sender' key? 回答1: You cannot turn off SRTP. It is part of the standard and it will probably never be

PJSUA: verbosity of debug information on the application console

由 烈酒焚心 提交于 2020-05-31 05:23:48
问题 If I create an application using PJSUA, then after the pjsua_create() system call, a huge amount of debugging information falls onto the console of my application. This is convenient at the development stage, but after it interferes with the work with programm. How can you predefine verbosity level of this debugging information? So that when pjsua_create() is called, it is already set. Thank you for the informative answers. 回答1: You can do this by calling pj_log_set_level(int level) (link)

ImportError: No module named sip (python2.7 PyQt4)

由 ＊爱你＆永不变心＊ 提交于 2020-05-29 11:09:07
问题 I'm currently using Ubuntu 18.04 LTS. I'm trying to install a program that need PyQt4 and QtWebKit, so a manual installation is necessary as QtWebKit have been excluded from PyQt4. I downloaded sip 4.19.12 (with 4.19.14 installation of PyQt4 fails) and PyQt4 4.12.13 I ran a virtualenv, made sure it was working as intended and tried installing sip, which works: $ python configure.py $ make $ make install Then I proceed with the same with PyQt4, with no errors. When I try to run my program .py,

ImportError: No module named sip (python2.7 PyQt4)

由 那年仲夏 提交于 2020-05-29 11:06:59
问题 I'm currently using Ubuntu 18.04 LTS. I'm trying to install a program that need PyQt4 and QtWebKit, so a manual installation is necessary as QtWebKit have been excluded from PyQt4. I downloaded sip 4.19.12 (with 4.19.14 installation of PyQt4 fails) and PyQt4 4.12.13 I ran a virtualenv, made sure it was working as intended and tried installing sip, which works: $ python configure.py $ make $ make install Then I proceed with the same with PyQt4, with no errors. When I try to run my program .py,

sip making a call - 401 Unauthorized

由 大憨熊 提交于 2020-05-29 09:41:32
问题 I am working on a sip client. I'm monitoring with wireshark the sip packets. Register with the sip server works fine When making a call I have this: Client - INVITE message Server - 401 UNAUTHORIZED Client - INVITE message Server - 403 Forbidden I do not have access to the server. What could go wrong? Why can't I make a call? What is with that 401 and than 403 if registered worked ok? 回答1: That is noramal. Server ask for authorization(401). After that it say authorization incorrect. That is

sip making a call - 401 Unauthorized

由 南笙酒味 提交于 2020-05-29 09:41:13
问题 I am working on a sip client. I'm monitoring with wireshark the sip packets. Register with the sip server works fine When making a call I have this: Client - INVITE message Server - 401 UNAUTHORIZED Client - INVITE message Server - 403 Forbidden I do not have access to the server. What could go wrong? Why can't I make a call? What is with that 401 and than 403 if registered worked ok? 回答1: That is noramal. Server ask for authorization(401). After that it say authorization incorrect. That is

sip making a call - 401 Unauthorized

由 て烟熏妆下的殇ゞ 提交于 2020-05-29 09:41:12
问题 I am working on a sip client. I'm monitoring with wireshark the sip packets. Register with the sip server works fine When making a call I have this: Client - INVITE message Server - 401 UNAUTHORIZED Client - INVITE message Server - 403 Forbidden I do not have access to the server. What could go wrong? Why can't I make a call? What is with that 401 and than 403 if registered worked ok? 回答1: That is noramal. Server ask for authorization(401). After that it say authorization incorrect. That is

FastJSON的0day漏洞报告

由 你。 提交于 2020-05-09 22:14:15
一、问题背景 fastjson是阿里巴巴的开源JSON解析库，它可以解析JSON格式的字符串，支持将Java Bean序列化为JSON字符串，也可以从JSON字符串反序列化到JavaBean，由于具有执行效率高的特点，应用范围很广 2019年6月22日，阿里云云盾应急响应中心监测到FastJSON存在0day漏洞，攻击者可以利用该漏洞绕过黑名单策略进行远程代码执行 关于fastjson javaweb框架的0day漏洞情报，由于fastjson在进行实例化对象时没有对输入数据进行严格限制，攻击者只要精心构造json数据，即可造成远程代码执行，截止到发稿日，关于该漏洞的利用方式暂未公开，请相关用户及时进行防护。 2019年6月22日，阿里云云盾应急响应中心监测到FastJSON存在0day漏洞，攻击者可以利用该漏洞绕过黑名单策略进行远程代码执行。 1.1、漏洞名称 FastJSON远程代码执行0day漏洞 1.2、漏洞描述 利用该0day漏洞，恶意攻击者可以构造攻击请求绕过FastJSON的黑名单策略。例如，攻击者通过精心构造的请求，远程让服务端执行指定命令（以下示例中成功运行计算器程序）。 1.3、影响范围 FastJSON 1.2.30及以下版本 FastJSON 1.2.41至1.2.45版本 1.4、官方解决方案 升级至FastJSON最新版本，建议升级至1.2.58版本。
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