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If SQL Server Anywhere is started via "dbsrv12.exe -z -x tcpip " it opens a Serverport.

Default is 2638.

If more then one database is opened, the other ones are getting "some" serverports.

How could I say: open the server on "this" port

In help of SQL anywhere I find: TCP/IP: Broadcast, BroadcastListener, DoBroadcast, Host, LDAP, LocalOnly, MyIP, ReceiveBufferSize, SendBufferSize, ServerPort, TDS, Timeout

Using dbsrv12.exe -z -x tcpip ServerPort 54382 gives 2 errors, that ServerPort and 54382 are not a file. states that:

You can specify a single port number, or a combination of individual port numbers and ranges of port numbers. When you specify a list and/or range of port numbers, the database server attempts to bind to all specified port numbers. For example: (port=1234)

Default TCP/IP 2638

Therfore i tried dbsrv12.exe -z -x tcpip (port=54382) and dbsrv12.exe -z -x tcpip 54382, both giving the same error as above.

In which way do I misunderstand the help?

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up vote 1 down vote accepted

The proper syntax is:

dbsrv12.exe -z -x tcpip(ServerPort=54382)

No space between tcpip and the (. You can use "Port=" instead of "ServerPort=" if you want.

share|improve this answer
thx, was just overlooking the space. this made the batch a lot easier which fires up a few databases (try to synchronize them by code) – Offler Nov 29 '12 at 16:35

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