The 'RESET' message is most likely somebody hitting the 'stop' button in the
browser. The 'connection timeout' is Tomcat hanging up the line when it
didn't get another request from Apache within 20sec of the last one (which
is what you told it to do :).

"Kommineni, Sateesh (GE Consumer & Industrial)" <sateesh.kommineni@(protected)>
wrote in message
news:F001DE493B12B74BB135AD692FC2BE36087AD63D@(protected)...
Hi All,

  I am trying to deploy a Web Application in my Tomcat and got the
following Error in the Log Files..

  Apr 20, 2004 11:05:04 AM org.apache.jk.server.JkCoyoteHandler action
INFO: RESET
Apr 20, 2004 11:05:04 AM org.apache.jk.server.JkCoyoteHandler action
INFO: RESET
Apr 20, 2004 11:05:14 AM org.apache.jk.common.ChannelSocket
processConnection
INFO: connection timeout reached

  What does this mean ??

  I have the following entries in my server.xml.

<Connector className="org.apache.coyote.tomcat4.CoyoteConnector"
acceptCount="10" bufferSize="2048" connectionTimeout="20000" debug="0"
enableLookups="true" maxProcessors="75" minProcessors="5"
address="3.130.164.201" port="12312"
protocolHandlerClassName="org.apache.coyote.http11.Http11Protocol"
proxyPort="0" redirectPort="8443" scheme="http" secure="false"
tcpNoDelay="true" useURIValidationHack="false">
<Factory className="org.apache.catalina.net.DefaultServerSocketFactory"/>
</Connector>

<Connector className="org.apache.coyote.tomcat4.CoyoteConnector"
acceptCount="50" bufferSize="2048" connectionTimeout="10000" debug="0"
enableLookups="false" maxProcessors="75" minProcessors="20"
address="3.130.164.201" port="12313"
protocolHandlerClassName="org.apache.jk.server.JkCoyoteHandler"
proxyPort="0" redirectPort="8443" scheme="ajp13" secure="false"
tcpNoDelay="true" useURIValidationHack="false">
Factory className="org.apache.catalina.net.DefaultServerSocketFactory"/>
</Connector>

 what is the significance of ajp13 scheme ??

Pls let me know.. Thanks a lot..

-Sateesh




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