How to pass syslog message without additional timestamp

Tony Reinke

I am trying to forward an exact duplicate of the syslog message. I have the route going thru passthru to not change the data. But I am getting the timestamp and original message host added to the message.

Thoughts?

Source: Syslog on 514/tcp/udp.

Pipeline: passthru

Source Data:
<164>:Jul 23 12:34:15 CDT: %ASA-auth-4-987654: Group = 1234567890, Username = efghijk, IP = aaa.bbb.ccc.ddd, Session disconnected. Session Type: LAN-to-LAN, Duration: 0h:30m:29s, Bytes xmt: 0, Bytes rcv: 0, Reason: Idle Timeout

Destination Data:
<164>Jul 23 12:34:15 lll.mmm.nnn.ppp :Jul 23 12:34:15 CDT: %ASA-auth-4-987654: Group = 1234567890, Username = efghijk, IP = aaa.bbb.ccc.ddd, Session disconnected. Session Type: LAN-to-LAN, Duration: 0h:30m:29s, Bytes xmt: 0, Bytes rcv: 0, Reason: Idle Timeout

Tony Reinke

Thanks to @Anson VanDoren and Oliver. The answer is to use an eval command with __syslogout and _raw.

Name = __syslogout
Value Expression = _raw

Doing this, allowed the syslog message to correct flow to the destination without adding the hop information.

Thanks to @Ralph No. for a link to the correct documentation.
https://docs.cribl.io/stream/destinations-syslog/#approaches

"The section about Output structure helps. It also has details on __syslogout vs _raw vs message..."

Tony Reinke

Thanks to @Anson VanDoren and Oliver. The answer is to use an eval command with __syslogout and _raw.

Name = __syslogout
Value Expression = _raw

Doing this, allowed the syslog message to correct flow to the destination without adding the hop information.

Thanks to @Ralph No. for a link to the correct documentation.
https://docs.cribl.io/stream/destinations-syslog/#approaches

"The section about Output structure helps. It also has details on __syslogout vs _raw vs message..."