Solved

Cursor out of scope

Posted on 2004-04-30
6
1,724 Views
Last Modified: 2008-02-01
FUNCTION  match_cit_feature
     (seq_id_in IN gcg_bioseq.seq_id%TYPE,
      pub_id_in IN gcg_publication.pub_id%TYPE)
  RETURN gcg_feature.feat_id%TYPE
  is
cursor c_pub is  select feat_id ,pub_id  from gcg_feature_pub
  where feat_id in (select feat_id from gcg_feature where seq_id = seq_id_in  );
  local_feat_id number(15);
  local_pub_id number(15);
  Begin
For i in  c_pub loop
if c_pub.pub_id = pub_id_in then
return c_pub.feat_id;
else
return null;
end if;
end loop;
  end match_cit_feature;
For the above set of code i get the error
Errors for PACKAGE BODY GCGFEAT_TEST:

LINE/COL ERROR
-------- -----------------------------------------------------------------
14/1     PL/SQL: Statement ignored
14/10    PLS-00225: subprogram or cursor 'C_PUB' reference is out of scope
What does out of scope error mean
0
Comment
Question by:nandini22
[X]
Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
6 Comments
 
LVL 44

Expert Comment

by:Arthur_Wood
ID: 10958422
the real cause of the error 14/10  is the first error 14/1  which suggests that there is a problem in your FUNCTION declaration.

I would suggest that you should check the PL/SQL syntax for declaring a FUNCTION as a 'Stored Procedure'. It has been about 7 yeears since I used ORACLE, so I can't give you the corect syntax, but my guess is that there is an error in the FUNCTION declaration.  

The 14/10 error is then due to the fact that you cannot declare a cursor outside of the definition of either a valid FUNCTION or a valid STORED PROCEDURE.

AW
0
 
LVL 22

Accepted Solution

by:
Helena Marková earned 125 total points
ID: 10958611
I think that this will work:

For rec in  c_pub loop
if rec.pub_id = pub_id_in then
 return rec.feat_id;
else
 return null;
end if;
end loop;
0
 

Author Comment

by:nandini22
ID: 10959194
Thanks Henka it works ,but I have another problem that when there are no rows selected by the cursor the exception to return a null is not being raised.
0
 
LVL 2

Assisted Solution

by:Elena-S
Elena-S earned 125 total points
ID: 10960695
It is not considered an exception to return null from a function.  If you would like to treat it as exception, you need to raise exception explicitly.  For example,

if some_value is null then
   raise_application_error(-20000, 'Your error message goes here');
end if;

You can find detailed description of this procedure in Oracle help available on-line.
0

Featured Post

On Demand Webinar: Networking for the Cloud Era

Did you know SD-WANs can improve network connectivity? Check out this webinar to learn how an SD-WAN simplified, one-click tool can help you migrate and manage data in the cloud.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Recently I was talking with Tim Sharp, one of my colleagues from our Technical Account Manager team about MongoDB’s scalability. While doing some quick training with some of the Percona team, Tim brought something to my attention...
In this series, we will discuss common questions received as a database Solutions Engineer at Percona. In this role, we speak with a wide array of MySQL and MongoDB users responsible for both extremely large and complex environments to smaller singl…
Polish reports in Access so they look terrific. Take yourself to another level. Equations, Back Color, Alternate Back Color. Write easy VBA Code. Tighten space to use less pages. Launch report from a menu, considering criteria only when it is filled…
This is a high-level webinar that covers the history of enterprise open source database use. It addresses both the advantages companies see in using open source database technologies, as well as the fears and reservations they might have. In this…

627 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question