on 10-01-2004 11:32 PM
Hi,
I am having some problems to install this versions: NWVC60_0-20000504
About my landscape:
I am using two boxes machines: one of them where is installed MSSQL 2000 SP3, and the other one E.P 6.0 SP2 Patch 4 and IIS 5.0.
I read Installation Guide 1.2 during the step 3.4 Installing Visual Composer Storyboard: I run setup.exe the system asked me:
HOST NAME:
DEFAULT TCP/IP PORT:
AND DBNAME
After entering of this paremeters appear a message saying:
"Visual Composer Installation requires SQL200 SERVER indicate appropiate server"
This is MSQLServer 2000 and I have installed SP3 (I don't understand this message).
In order to proof this connection I have configured an ODBC connection and it worked fine. (I use the same database, user and password).
Could you tell me what I am doing wrong ?
Here, I have a doubt:
Must I manualy create this D.B or the system create it ?
Thanks a lot,
Patricio Garcia.
Hi all,
to solve this install the client tools from the sql Server 2000 cd on the VC Server System.
Best regards
Thorsten Stracke
You must be a registered user to add a comment. If you've already registered, sign in. Otherwise, register and sign in.
Hi,
You don't have to manually create the database, it is done by the installer. But the installer stops on your case when testing the DB version. Try running the following SQL phrase used by the installer:
use master ;declare @dbversion varchar(150) ;select @dbversion = substring(@@version, 23,5);print '[Server version]'print @dbversion
It should result with the following:
[Server version]
2000
It also writes it into a temp file in a temp dir which might have failed for some reason (full administrative rights when installing?).
Regards,
Yaniv.
You must be a registered user to add a comment. If you've already registered, sign in. Otherwise, register and sign in.
User | Count |
---|---|
87 | |
10 | |
10 | |
10 | |
7 | |
6 | |
6 | |
5 | |
5 | |
4 |
You must be a registered user to add a comment. If you've already registered, sign in. Otherwise, register and sign in.