Thread: Query tool not working
Hi,<br /><br />I am new to postgeSQL and have recently installed version 8.1 on my local machine.<br /><br />I am runningWindows XP Pro and am using pgAdmin III version 1.4.1<br /><br />I created a database with all the required tablesand columns. I have also created all the primary keys and foreign keys and can view/ created columns and tables andsequences. At the moment there is no data within the tables. <br /><br />The problem is, when I use the query tool todo a select all statement, It comes back with an error message saying the the table could not be found. I have tried iton most of the tables, but get the same error message. I am spelling the name of the table correctly. <br /><br />Anyideas on how to fix this? Would it have something to do with permissions? <br />
From: pgadmin-support-owner@postgresql.org [mailto:pgadmin-support-owner@postgresql.org] On Behalf Of Ryan Dobbs
Sent: 03 March 2006 03:20
To: pgadmin-support@postgresql.org
Subject: [pgadmin-support] Query tool not workingHi,
I am new to postgeSQL and have recently installed version 8.1 on my local machine.
I am running Windows XP Pro and am using pgAdmin III version 1.4.1
I created a database with all the required tables and columns. I have also created all the primary keys and foreign keys and can view/ created columns and tables and sequences. At the moment there is no data within the tables.
The problem is, when I use the query tool to do a select all statement, It comes back with an error message saying the the table could not be found. I have tried it on most of the tables, but get the same error message. I am spelling the name of the table correctly.
Any ideas on how to fix this? Would it have something to do with permissions?
Unlikely - it would say so if that were the case.
I suspect you have either used upper case characters in your table name, and then not quoted it in your query - eg.
SELECT * FROM "FoO"
or, created the table in a schema not in the current search path, and not qualified the name, such as:
SELECT * FROM myschema.foo
Regards, Dave.