Issue 13736

Summary:	ODBC is not working in Linux
Product:	Base	Reporter:	pueyo <pueyo>
Component:	code	Assignee:	Frank Schönheit <frank.schoenheit>
Status:	CLOSED FIXED	QA Contact:	issues@dba <issues>
Severity:	Trivial
Priority:	P3	CC:	issues
Version:	OOo 1.1 Beta
Target Milestone:	---
Hardware:	PC
OS:	Linux, all
Issue Type:	DEFECT	Latest Confirmation in:	---
Developer Difficulty:	---

Description pueyo 2003-04-23 09:49:52 UTC

The ODBC enviroment (unixODBC) is not working with 1.1Beta in Linux (but it
works with win32). The same configuration worked (works) perfectly with OO.org
1.0.2. When I try to press the ODBC DataSource buttom an error message says that
the libodbc.so library is corrupted or something.

Since I am testing, I tried to use the libraries from OO 1.0.2 and copied them
to the 1.1Beta location. Same error.

Then I tried to use the general libraries I have in /usr/lib/ (by coping it to
OO.org-1.1Beta-installation/program/ .. I know ... but I am testing) and
(surprise!) now the DataSource dialog does work and I can select one of my ODBC
registered DataSource. Well, they do not work either because OO cannot make the
connection and says that the driver is not avalaible... When you choose ODBC as
the "database type", you get this scaring message: "This kind of data source is
not supported in this plataform" at the bottom of the Data Sources Administrator
Dialog. Hey, I hope ODBC support comes back to THE platform before final release
of 1.1...

Comment 1 pueyo 2003-05-01 19:27:46 UTC

I reinstalled OO.org 1.1 Beta and made this symbolic link:

ln -s /usr/lib/libodbc.so   OO.orgInstallation/program/libodbc.so

and everything is working perfect now with MySql and Postgres, even 
the Drag and Drop bug I reported for OO.org 1.0.2 has been solved ... 
great!

Comment 2 Frank Schönheit 2003-05-02 08:22:15 UTC

closing as FIXED (claimed by reporter)
pueyo, thanks for updating us ...

Comment 3 Frank Schönheit 2003-05-02 08:22:24 UTC

closing

Comment 4 hans_werner67 2004-02-02 13:04:40 UTC

change subcomponent to 'none'