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From Suresh Srinivas <sur...@hortonworks.com>
Subject Re: ClientProtocol Version mismatch. (client = 69, server = 1)
Date Tue, 29 Jan 2013 20:55:10 GMT
Please take this up in CDH mailing list. Most likely you are using client
that is not from 2.0 release of Hadoop.


On Tue, Jan 29, 2013 at 12:33 PM, Kim Chew <kchew534@gmail.com> wrote:

> I am using CDH4 (2.0.0-mr1-cdh4.1.2) vm running on my mbp.
>
> I was trying to invoke a remote method in the ClientProtocol via RPC,
> however I am getting this exception.
>
> 2013-01-29 11:20:45,810 ERROR
> org.apache.hadoop.security.UserGroupInformation:
> PriviledgedActionException as:training (auth:SIMPLE)
> cause:org.apache.hadoop.ipc.RPC$VersionMismatch: Protocol
> org.apache.hadoop.hdfs.protocol.ClientProtocol version mismatch.
> (client = 69, server = 1)
> 2013-01-29 11:20:45,810 INFO org.apache.hadoop.ipc.Server: IPC Server
> handler 6 on 8020, call
> org.apache.hadoop.hdfs.protocol.ClientProtocol.getFileInfo from
> 192.168.140.1:50597: error: org.apache.hadoop.ipc.RPC$VersionMismatch:
> Protocol org.apache.hadoop.hdfs.protocol.ClientProtocol version
> mismatch. (client = 69, server = 1)
> org.apache.hadoop.ipc.RPC$VersionMismatch: Protocol
> org.apache.hadoop.hdfs.protocol.ClientProtocol version mismatch.
> (client = 69, server = 1)
>         at
> org.apache.hadoop.ipc.ProtobufRpcEngine$Server$ProtoBufRpcInvoker.getProtocolImpl(ProtobufRpcEngine.java:400)
>         at
> org.apache.hadoop.ipc.ProtobufRpcEngine$Server$ProtoBufRpcInvoker.call(ProtobufRpcEngine.java:435)
>         at org.apache.hadoop.ipc.RPC$Server.call(RPC.java:898)
>         at org.apache.hadoop.ipc.Server$Handler$1.run(Server.java:1693)
>         at org.apache.hadoop.ipc.Server$Handler$1.run(Server.java:1689)
>         at java.security.AccessController.doPrivileged(Native Method)
>         at javax.security.auth.Subject.doAs(Subject.java:396)
>         at
> org.apache.hadoop.security.UserGroupInformation.doAs(UserGroupInformation.java:1332)
>         at org.apache.hadoop.ipc.Server$Handler.run(Server.java:1687)
>
> I could understand if the Server's ClientProtocol has version number
> "60" or something else, but how could it has a version number of "1"?
>
> Thanks.
>
> Kim
>



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