Issue with execution of PLSQL block in Shell script

Question

I am trying to drop tables, which have a particular suffix(passed as argument $1), using shell script.

If a parent table is selected without its child tables being dropped, I am bypassing the parent table and increasing the counter in the exception block.

When I run this script in sql developer with $1 replaced with the proper value, it works. But when I run this shell script, it is getting stuck.

Could you please have a look and let me know, what am I missing in the shell script?

Code:

#!/bin/bash

cat <<ENDDROPNEWTABS >dropnewtabs.sql
set pagesize 100

DECLARE

t_cnt NUMBER;
CURSOR C001
IS
SELECT table_name FROM user_tables WHERE table_name LIKE '%$1%';

BEGIN
BEGIN SELECT COUNT(*) INTO t_cnt FROM user_tables WHERE table_name LIKE '%$1%'; 
END;

WHILE(t_cnt > 0) LOOP
FOR i IN C001 LOOP
BEGIN EXECUTE IMMEDIATE 'DROP TABLE '||i.table_name;
EXCEPTION
WHEN OTHERS THEN
t_cnt := t_cnt+1;
NULL;
END;      
t_cnt := t_cnt-1;
END LOOP;
END LOOP;

END;

exit
ENDDROPNEWTABS

echo "Dropping the tables created for this task..."

sqlplus -s usn/pwd@sid @dropnewtabs.sql >tablesDropped.txt

#END

Answer 1

You are missing a / after the END; of your anonymous block, so it will never execute it, and the exit will be seen as part of the previous command. The / is very roughly analogous to 'run' in SQL Developer.

...
END LOOP;

END;
/

exit
ENDDROPNEWTABS

(You don't need the BEGIN/END around the SELECT, or the NULL in the exception handler, but those won't break anything; it's also not a good idea to squash all possible exceptions silently, just look for the one you're expecting to see. And personally I find it easier to follow with some indentation).