I want to get the number of sessions in Oracle using the SQL query:
SELECT value FROM v$parameter WHERE name = 'sessions'
But I get this error:
Error starting at line 1 in command:
SELECT value FROM v$parameter WHERE name = 'sessions'
Error at Command Line:1 Column:18
Error report:
SQL Error: ORA-00942: table or view does not exist
00942. 00000 - "table or view does not exist"
*Cause:
*Action:
Maybe the Oracle user that I use is not privileged?
Generally the better approach is to use a procedure and grant the necessary privileges to this procedure. However if you want use SQL directly, you can grant SELECT_CATALOG_ROLE or SELECT ANY DICTIONARY to the user.
Probably. To grant the rights, you need to use the table name as V_$PARAMETER. It comes from some restriction when granting rights on dynamic views.
If you want to use SQL directly (referring to the second option in the accepted answer)
As of Feb 2023, using Oracle version 19, this works...
Connect as SYSTEM and run
grant SELECT ANY DICTIONARY to <user>;
But SELECT_CATALOG_ROLE didn't work for me...
grant SELECT_CATALOG_ROLE to ...;
There are documented differences between the two here:
http://www.petefinnigan.com/weblog/archives/00001461.htm
This is where the author gives the following info and warning:
Well, SELECT_CATALOG_ROLE allows access to some things Oracle deemed not allowed by SELECT ANY DICTIONARY so we need to be careful of granting this role on these grounds. BUT, the overwhelming issue for me is that SELECT_CATALOG_ROLE gives access to 4539 objects and SELECT ANY DICTIONARY gives access to 6228 objects (both numbers in 18c XE)
I am not sure why Oracle do not publish the full list of exclusions in SELECT ANY DICTIONARY but they do publish all of the main tables. We can easily find out anyway. For me, i want to know what does SELECT ANY DICTIONARY really mean. I want to know what i am actually granting if I give out that privilege; well it means access to 6228 tables and views in 18cXE
Both of these rights should not be used; they are a sledgehammer to crack a peanut. If someone needs access to V$SESSION or V$DATABASE and there is a legitimate reason to have that access then grant access on the individual views not SELECT ANY DICTIONARY or SELECT_CATALOG_ROLE.
using the privileges: - select any table, alter any table when running the grant as SYS with SYSDBA in Oracle 12c solved the issue for me.
Related
I have created a new view named CONS_INTERRUPTED_DATA for the main user hfdora and the view has been created successfully. But when I am trying to create the same view for another user (cis) of the same database after giving all the privileges to the user (cis) I am getting the below error,
*oms_consumer
ERROR at line 13:
ORA-00942: table or view does not exist
Both the user hfdora and cis are the part of same database and this oms_consumer table is present at the database
I have granted the following privileges for the user cis before creating the view
grant select on energization_info to cis;
grant select on trigger_info to cis;
grant select on oms_source to cis;
grant select on oms_consumer to cis;
grant connect,resource,dba to cis;
My sql query to create the view,
>CREATE OR REPLACE VIEW CONS_INTERRUPTED_DATA AS
SELECT
trigger_info_A.b1 AS FDR_RMU_OFF_B1, trigger_info_A.b2 AS FDR_RMU_OFF_B2,
trigger_info_A.B3TEXT AS FDR_RMU_OFF_B3TEXT, trigger_info_A.elem AS FDR_RMU_OFF_ELEM,
trigger_info_B.b1 AS FDR_RMU_RESTORE_B1, trigger_info_B.b2 AS FDR_RMU_RESTORE_B2,
trigger_info_B.B3TEXT AS FDR_RMU_RESTORE_B3TEXT,
trigger_info_B.elem AS FDR_RMU_RESTORE_ELEM,
oms_consumer.consumer_code, energization_info.b1 AS AFFECTED_B1,
energization_info.b2 AS AFFECTED_B2, energization_info.b3text AS AFFECTED_B3TEXT,
to_char(energization_info.deenergized_date, 'DD-MM-YYYY Hh24:MI:SS') AS DEENERGIZED_DATE,
to_char(energization_info.energized_date, 'DD-MM-YYYY Hh24:MI:SS') AS ENERGIZED_DATE,
trigger_info_A.comments AS KEY
FROM
energization_info,
trigger_info trigger_info_A,
trigger_info trigger_info_B,
oms_consumer
WHERE
(energization_info.trigger_number = trigger_info_A.trigger_number)
AND (energization_info.ENERGIZED_TRIGGER_NUMBER = trigger_info_B.trigger_number)
AND (energization_info.b1 = oms_consumer.B1NAME
AND energization_info.b2 = oms_consumer.B2NAME
AND energization_info.b3 = oms_consumer.B3NAME)
WITH READ ONLY;
The first step in diagnosing a problem when creating a view is to try the select part on its own. In this case you would still get the ORA-00942 error, but the problem is now just a query and access issue and not to do with the view specifically.
When you get ORA-00942: table or view does not exist, it's because either:
The table or view name that you typed really doesn't exist.
Check the spelling - maybe there is a typo.
Are you connected to a database where it exists? Perhaps you are on a test system that doesn't have it.
Query dba_objects to see whether the table exists in another schema. (If you don't have privileges to query dba_objects, all_objects lists everything you have permission to view, which may be some help.)
It really does exist, but it's in another schema.
In that case, there are two possible issues:
You don't have permission to query it. The table's owner needs to grant read on xyz (substitute the actual table name for xyz) to either
you
public (if you want everyone to be able to see the data, not always advisable)
a role that you have (but roles aren't used by stored PL/SQL or views, though, so it's possible that you can query a table in another schema thanks to a role that you have, but still not be able to create a view or a procedure that uses it.)
You need to specify the schema. Say you want to query the REGIONS table in HR but you are connected as SCOTT. If you just select * from regions it will look for SCOTT.REGIONS, which doesn't exist. To fix that, do one of the following:
use hr.regions explicitly in your query.
in your schema, create or replace synonym regions for hr.regions;
Now whenever you refer to regions, the database will automatically redirect to hr.regions.
in any schema with permission to create public synonyms:
create or replace public synonym regions for hr.regions;
Now everyone connecting to the database will have any references to regions redirected to hr.regions, which isn't always a good idea, but it's one option anyway.
alter session set current_schema = hr;
Now the default schema for resolving names of objects is HR and not the one you logged into. For applications that always log in as a different user than the one that owns the tables, you can create an after logon trigger so this is always set. Then they can just refer to regions etc without needing to specify any schema and without any synonyms.
My issue has been resolved. :-)
I have made the following changes,
FROM
hfdora.energization_info,
hfdora.trigger_info trigger_info_A,
hfdora.trigger_info trigger_info_B,
hfdora.oms_consumer
Now the same view is created for the user cis.
I am using oracle 10g express edition. I ran the command in sql command line and it says -- "ORA-00942: table or view does not exist"
PS: Trying to access and increase the number of processes. Cause I am facing this problem -- How to solve ORA-12516 error?
You're probably running this command as a non-privileged user. show parameter is just a fancy wrapper for select ... from v$parameter. You need SELECT privileges on this view:
grant select on v_$parameter to <username>;
(please note the _ in the view name - you cannot directly grant privileges on v$ views, you have to grant privileges on the underlying objects instead).
Of course, the simplest approach is to run show parameter as a DBA user.
I want to execute a query that selects data from a different schema than the one specified in the DB connection (same Oracle server, same database, different schema)
I have an python app talking to an Oracle server. It opens a connection to database (server/schema) A, and executes select queries to tables inside that database.
I've tried the following :
select ....
from pct.pi_int, pct.pi_ma, pct.pi_es
where ...
But I get:
ORA-00942: table or view does not exist
I've also tried surrounding the schema name with brackets:
from [PCT].pi_int, [PCT].pi_ma, [PCAT].pi_es
I get:
ORA-00903: invalid table name
The queries are executed using the cx_Oracle python module from inside a Django app.
Can this be done or should I make a new db connection?
Does the user that you are using to connect to the database (user A in this example) have SELECT access on the objects in the PCT schema? Assuming that A does not have this access, you would get the "table or view does not exist" error.
Most likely, you need your DBA to grant user A access to whatever tables in the PCT schema that you need. Something like
GRANT SELECT ON pct.pi_int
TO a;
Once that is done, you should be able to refer to the objects in the PCT schema using the syntax pct.pi_int as you demonstrated initially in your question. The bracket syntax approach will not work.
In addition to grants, you can try creating synonyms. It will avoid the need for specifying the table owner schema every time.
From the connecting schema:
CREATE SYNONYM pi_int FOR pct.pi_int;
Then you can query pi_int as:
SELECT * FROM pi_int;
Depending on the schema/account you are using to connect to the database, I would suspect you are missing a grant to the account you are using to connect to the database.
Connect as PCT account in the database, then grant the account you are using select access for the table.
grant select on pi_int to Account_used_to_connect
My application has few tables on Oracle where user XYZ is the schema owner. Tables has been created using XYZ. And I would like to have ABCUSER to have CRUD rights on these tables. I have given the access via GRANT ALL ON TABLEABC to ABCUSER and grant is succeeded.
But when this user(ABCUSER) tries to query the DB (select * from TABLEABC) it doesn't seem to work. I get error message
ORA-00942: table or view does not exist
00942. 00000 - "table or view does not exist"
*Cause:
*Action:
Error at Line: 1 Column: 14
Could you please tell what am I missing ?
User ABCUSER has privileges on the table but doesn't own it. So you need to include the schema in the query:
select * from XYZ.TABLEABC
/
If you don't want to hardcode the schema name in your programs your user can build a synonym:
create synonym TABLEABC for XYZ.TABLEABC
/
Then the original query will work for ABCUSER.
Note that ABCUSER will require the CREATE SYNONYM privilege.
As APC says, you are missing a SYNONYM.
You might want either a PRIVATE or PUBLIC synonym depending upon who you want to be able to view the table.
A good description of the various types and their uses is here:
http://www.orafaq.com/wiki/Synonym
I am using insert statement and trying to insert data into the database table. I am using stored procedures.
But I am getting this error while doing so.
Message: ORA-00942: table or view does
not exist ORA-06512
I checked if the tables/stored procedures are present or not and everything is in place. Also there is no typo in table names or in sp. If I run the part of SP from query editor it works fine but when I execute the entire SP it throws an error.
I tried the steps provided by Stephen but since I have logged in with the same user/owner when I run Grant command it gives me an error saying 'Cannot Grant/revoke on own'.
One more addition to this. I have a stored procedure SP1 in which I am using a select statement as
Select a from table_name where condition;
When I execute this seperately, it returns me some results. But when I execute sp it gives an error at the same line where it is written.
Can anyone help me out to resolve this issue. I am using SQL +.
Thanks in advance
Vijay
Justin's answer is correct but let me expand a bit.
Everyone who said that the table doesn't exist didn't read your whole post. Since you are able to:
If I run the part of SP from query editor it works fine
Obviously the table is there.
Obviously you have some access to it. Otherwise this wouldn't work when it clearly does.
but when I execute the entire SP it throws an error.
This is because Oracle distinguishes between permissions granted directly and those granted via a role.
Say I do this:
Create Table TABLE_A
Create Role READ_ONLY
Grant Select on TABLE_A to READ_ONLY
Grant READ_ONLY to VIJAY
In a SQL Window/prompt you could query that table without issue. So now you need to create a view
Create VIJAY.VIEW_A as SELECT * FROM TABLE_A
You'll get the error that TABLE_A does exist. Because a view is compiled, like a procedure it runs without any roles. Since it runs without the READ_ONLY role, it's blind to the fact that TABLE_A exists. Now what I need to do is
Grant Select on TABLE_A to VIJAY.
Now that you have a direct permission, you can compile a view or procedure/package that uses that table.
Does the table exist in the schema where the stored procedure exists? If not, the simplest explanation is that the owner of your procedure has been granted access to the table via a role not via a direct grant. A definer's rights stored procedure needs to have direct access to the objects it accesses. A quick way to test this is to disable roles for the session, i.e.
SQL> set role none;
SQL> <<execute your query>>
If that generates the error, the problem is the lack of a direct grant.
In Oracle you can choose if the stored procedure is executed with the rights of the invoker or the definer: http://download.oracle.com/docs/cd/E11882_01/appdev.112/e17126/subprograms.htm#i18574
Check if the AUTHID property of the stored procedure is correct and if the resulting user has appropriate permissions.
Well, put very simply, the table that you are trying to insert data into does not exist in the database you are connected to. You need to check both those things (i.e. what are you connected to, and is the table there and accessible for the user context you are using).
As Joe Stefanelli said .. there are a lot of possibilities for the error being shown here.
Check whether:
You are connecting to the correct Oracle Instance.
You have permissions to query or perform processing on table that you are referencing in your query.
There is a difference between ordinary select statements and procedures. Procedures in oracle do not respect the roles assigned to a user; rather the permission needs to be explicitly granted to the user. For more information read the following linkORA-00942