5

我被这个问题困住了，我无法解决它。

代码：

 ConnectionConfiguration config = new ConnectionConfiguration(host, port);
        config.setDebuggerEnabled(true);
        config.setSASLAuthenticationEnabled(true);
        config.setSecurityMode(ConnectionConfiguration.SecurityMode.enabled);

        XMPPConnection connection = new XMPPConnection(config);
try{
            connection.connect();
            connection.login(login, password);
            //auth.authenticate(login, password, host);
        }
        catch (XMPPException exc){
            exc.printStackTrace();
        }

完整的堆栈跟踪：

/Library/Java/JavaVirtualMachines/1.6.0_35-b10-428.jdk/Contents/Home/bin/java -Didea.launcher.port=7544 "-Didea.launcher.bin.path=/Applications/IntelliJ IDEA 11 CE.app/bin" -Dfile.encoding=UTF-8 -classpath "/Library/Java/JavaVirtualMachines/1.6.0_35-b10-428.jdk/Contents/Home/lib/deploy.jar:/Library/Java/JavaVirtualMachines/1.6.0_35-b10-428.jdk/Contents/Home/lib/dt.jar:/Library/Java/JavaVirtualMachines/1.6.0_35-b10-428.jdk/Contents/Home/lib/javaws.jar:/Library/Java/JavaVirtualMachines/1.6.0_35-b10-428.jdk/Contents/Home/lib/jce.jar:/Library/Java/JavaVirtualMachines/1.6.0_35-b10-428.jdk/Contents/Home/lib/jconsole.jar:/Library/Java/JavaVirtualMachines/1.6.0_35-b10-428.jdk/Contents/Home/lib/management-agent.jar:/Library/Java/JavaVirtualMachines/1.6.0_35-b10-428.jdk/Contents/Home/lib/plugin.jar:/Library/Java/JavaVirtualMachines/1.6.0_35-b10-428.jdk/Contents/Home/lib/sa-jdi.jar:/Library/Java/JavaVirtualMachines/1.6.0_35-b10-428.jdk/Contents/Classes/charsets.jar:/Library/Java/JavaVirtualMachines/1.6.0_35-b10-428.jdk/Contents/Classes/classes.jar:/Library/Java/JavaVirtualMachines/1.6.0_35-b10-428.jdk/Contents/Classes/jsse.jar:/Library/Java/JavaVirtualMachines/1.6.0_35-b10-428.jdk/Contents/Classes/ui.jar:/Library/Java/JavaVirtualMachines/1.6.0_35-b10-428.jdk/Contents/Home/lib/ext/apple_provider.jar:/Library/Java/JavaVirtualMachines/1.6.0_35-b10-428.jdk/Contents/Home/lib/ext/dnsns.jar:/Library/Java/JavaVirtualMachines/1.6.0_35-b10-428.jdk/Contents/Home/lib/ext/localedata.jar:/Library/Java/JavaVirtualMachines/1.6.0_35-b10-428.jdk/Contents/Home/lib/ext/sunjce_provider.jar:/Library/Java/JavaVirtualMachines/1.6.0_35-b10-428.jdk/Contents/Home/lib/ext/sunpkcs11.jar:/Users/dmitry/IdeaProjects/smacker/target/classes:/Users/dmitry/IdeaProjects/smacker/lib/smack-3.2.1.jar:/Users/dmitry/IdeaProjects/smacker/lib/smackx-3.2.1.jar:/Applications/IntelliJ IDEA 11 CE.app/lib/idea_rt.jar" com.intellij.rt.execution.application.AppMain Main
Connection failed. No response from server.: 
    at org.jivesoftware.smack.PacketReader.startup(PacketReader.java:119)
    at org.jivesoftware.smack.XMPPConnection.initConnection(XMPPConnection.java:568)
    at org.jivesoftware.smack.XMPPConnection.connectUsingConfiguration(XMPPConnection.java:527)
    at org.jivesoftware.smack.XMPPConnection.connect(XMPPConnection.java:953)
    at SessionsHandler.openSession(SessionsHandler.java:62)
    at Main.main(Main.java:16)
    at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
    at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:39)
    at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:25)
    at java.lang.reflect.Method.invoke(Method.java:597)
    at com.intellij.rt.execution.application.AppMain.main(AppMain.java:120)

Smack 调试输出：

在：

<stream:stream to="xmpp.odnoklassniki.ru" xmlns="jabber:client" xmlns:stream="http://etherx.jabber.org/streams" version="1.0">
<starttls xmlns="urn:ietf:params:xml:ns:xmpp-tls"/>
<stream:stream to="odnoklassniki.ru" xmlns="jabber:client" xmlns:stream="http://etherx.jabber.org/streams" version="1.0">
</stream:stream>

出去：

<?xml version="1.0"?><stream:stream xmlns="jabber:client" xmlns:stream="http://etherx.jabber.org/streams" from="odnoklassniki.ru" version="1.0" xml:lang="en">
<stream:features><starttls xmlns='urn:ietf:params:xml:ns:xmpp-tls'/><mechanisms xmlns='urn:ietf:params:xml:ns:xmpp-sasl'><mechanism>X-ODKL-API</mechanism><mechanism>X-ODKL-BLOWFISH</mechanism><mechanism>PLAIN</mechanism></mechanisms><auth xmlns='http://jabber.org/features/iq-auth'/></stream:features>
<proceed xmlns='urn:ietf:params:xml:ns:xmpp-tls'/>
<?xml version="1.0"?><stream:stream xmlns="jabber:client" xmlns:stream="http://etherx.jabber.org/streams" from="odnoklassniki.ru" version="1.0" xml:lang="en">
<stream:features><mechanisms xmlns='urn:ietf:params:xml:ns:xmpp-sasl'><mechanism>X-ODKL-API</mechanism><mechanism>X-ODKL-BLOWFISH</mechanism><mechanism>PLAIN</mechanism></mechanisms><auth xmlns='http://jabber.org/features/iq-auth'/></stream:features>

远程登录：

dmitry@pro:~$ telnet xmpp.odnoklassniki.ru 5222
Trying 217.20.145.69...
Connected to xmpp.odnoklassniki.ru.
Escape character is '^]'.
4

4 回答 4

8

Smack 库 (3.2.2) 在第二个（在 starttls 之后）“stream:stream”数据包中需要 @id 属性。

此属性是可选的，并且 xmpp.odnoklassnki.ru 不发送它：

<stream:stream to="odnoklassniki.ru" xmlns="jabber:client" xmlns:stream="http://etherx.jabber.org/streams" version="1.0">

有两种选择可以解决这个问题：

  • 修补 PacketReader 以将某种“假”ID 分配给流中没有 @id 属性的连接：流数据包
  • 或要求 xmpp.odnoklassniki.ru 的管理员/开发人员将（假的或真实的）ID 属性添加到 stream:stream 数据包

这是 PacketReader:parsePackets() 方法的“修补”部分。请注意新的 connectionIdReceived 变量和对应的“if”条件：

// We found an opening stream. Record information about it, then notify
// the connectionID lock so that the packet reader startup can finish.
else if (parser.getName().equals("stream")) {
    // Ensure the correct jabber:client namespace is being used.
    if ("jabber:client".equals(parser.getNamespace(null))) {
        // Get the connection id.
        boolean connectionIdReceived = false;
        for (int i=0; i<parser.getAttributeCount(); i++) {
            if (parser.getAttributeName(i).equals("id")) {
                // Save the connectionID
                connectionID = parser.getAttributeValue(i);
                if (!"1.0".equals(parser.getAttributeValue("", "version"))) {
                    // Notify that a stream has been opened if the
                    // server is not XMPP 1.0 compliant otherwise make the
                    // notification after TLS has been negotiated or if TLS
                    // is not supported
                    releaseConnectionIDLock();
                }
                connectionIdReceived = true;
            }
            else if (parser.getAttributeName(i).equals("from")) {
                // Use the server name that the server says that it is.
                connection.config.setServiceName(parser.getAttributeValue(i));
            }
        }
        if (!connectionIdReceived) {
            connectionID = "<unspecified>";
            if (!"1.0".equals(parser.getAttributeValue("", "version"))) {
                // Notify that a stream has been opened if the
                // server is not XMPP 1.0 compliant otherwise make the
                // notification after TLS has been negotiated or if TLS
                // is not supported
                releaseConnectionIDLock();
            }
        }
    }
}
于 2012-12-07T14:37:47.663 回答
3

我看到了几种可能性：

  • 由于信任链或服务器提供的 *-name，您的库不接受您从服务器获取的证书：

主题：C=RU，L=莫斯科，O=Odnoklassniki Ltd，CN=*.odnoklassniki.ru

  • 在“安全”模式下，您的库不喜欢该服务器提供的任何身份验证机制，因为只有两个非标准机制（X-ODKL-API、X-ODKL-BLOWFISH）和两个更少-安全机制（PLAIN，http://jabber.org/features/iq-auth）。

如果我不得不猜测它会是第二个。这不是大多数客户端库作者不太可能遇到的错误情况。关闭安全模式，然后重试。

于 2012-11-12T04:07:35.997 回答
3

您的 XMPP 域似乎是错误的，这也可能是问题所在。您还将两个流标头发送到不同的域：

标题 1 到xmpp.odnoklassniki.ru：

<stream:stream to="xmpp.odnoklassniki.ru" xmlns="jabber:client" xmlns:stream="http://etherx.jabber.org/streams" version="1.0">

odnoklassniki.ru的标题 2 ：

<stream:stream to="odnoklassniki.ru" xmlns="jabber:client" xmlns:stream="http://etherx.jabber.org/streams" version="1.0">

服务器总是回复 .... from="odnoklassniki.ru" .... 所以odnoklassniki.ru应该是您正确的 XMPP 域。

于 2012-11-15T20:29:19.160 回答
1

尝试这个...

它对我有用...尝试也给资源名称...我面临同样的问题 gioving 资源名称解决了它

 ConnectionConfiguration config;
config = new ConnectionConfiguration(ip, port, "Smack");
config.setSASLAuthenticationEnabled(false);
config.setReconnectionAllowed(true);
config.setCompressionEnabled(false);
config.setDebuggerEnabled(true);
connection = new XMPPConnection(config);
connection.connect();
if (connection.isConnected()) {
    connection.login(loginName, password);
}
于 2012-11-17T10:07:02.777 回答