working late, i make typos.
the user i did create was called 'ifm' -- i was trying to change all the usernames to 'bar' to anonymize things.
so i forgot to change the last string when i pasted it to my previous email :-)

what i sent are the command i used in order.

if i shutdown the 'mserver5' , and restart it, i am able to login to my database once.
then it won't let me in anymore.

i created the 'voc' example, and have had no problem logging in and out
as use 'voc'.  so i'm inclined to think i am doing something wrong in either
creating the database, user or schema.

-kevin

On Apr 16, 2008, at 2:22 AM, Fabian Groffen wrote:

On 16-04-2008 01:52:22 -0400, kykim_lists@mac.com wrote:
I found this subject in the archives, and I am having the same problem
I am running on Mac OS X 10.4.11
I built MonetDB from source from MonetDB-Feb2008-SuperBall

I am new to MonetDB, so please forgive any ignorance..

Here's what I did:
> monetdb create foo # create the db
> mserver5 --dbname=foo --dbinit="include sql;" # start the db

> ./mclient -lsql
sql>CREATE USER "bar" WITH PASSWORD 'password' NAME 'Foobar User'  
SCHEMA "sys";
sql>CREATE SCHEMA "bar" AUTHORIZATION "bar";
sql>ALTER USER "bar" SET SCHEMA "bar";
> ./mclient -lsql -ubar
Password:password
sql>\q
> ./mclient -lsql -ubar
Password:password
!InvalidCredentialsException:checkCredentials:Invalid credentials  
for user 'ifm'

Two things:
1) I am able to login as user 'monetdb' without any problem
2) If i stop and start 'mserver5', I can login ONCE

I know that I probably should be using 'merovingian', but I'm still  
playing around.

Any pointers?  Anything I can do to help debug?

Well, the only thing I'm worried about here is that you mix two ways of
running a server.

You create the database using the `monetdb` tool, but start it manually.
While this in itself is no problem, it becomes a problem when
merovingian starts that database.  However, you mention you don't use
it.  Is the above the exact sequence of commands you perform to get a
database that refuses you?  I do notice that the error you get back
refers to a user "ifm", whereas the commandline argument suggests a user
"bar".

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