Sql Query on subqueries

Question

Suppose one table "T" is having single coloumn "ID" with the values like 1,2,3,4,5,6,8,9,10 (Note: there is no 7). Now I want to display only 3,5,7,8,9 with select statement. I can write like
SQL> select ID from T where ID in (3,5,7,8,9);
Now It will display only rows 3,5,8,9 but not 7.
My question is, how to display the row which does not exist (In this case 7) in my list.
Please help me on this.
Raghu N.
Edited by: user3270580 on May 3, 2010 3:00 AM

Answer 1

Hi,
user3270580 wrote:
Ex:
create table rag (nid number(2),sid number(3));
insert into rag values (1,100);
insert into rag values (2,100);
insert into rag values (3,100);
insert into rag values (5,100);
insert into rag values (6,100);
insert into rag values (8,100);
insert into rag values (9,100);
update rag set sid = 200 where nid in (2,3,4,5,7,8);
============
In this case, update statement will update only 4 rows (NIDs 2,3,5, and 8). We want to tell the customer that, update operatoin was not done for the rows (nid= 4 and nid=7), since they does not exist. Like wise we want to list all the non existing NIDs from list given by my customer.
In your early query I understand that, we should have a table which contains the list of rows(nids) for which we need to peroform update operation (correct me if i am wrong). But, as per my customer policy, I should not create any new table in the database.No. I said you need a table OR A RESULT SET . The two queries I posted used a result set generated in the query iteself. As you can see, no new table was created.
The solutions I posted earlier will give you the right results. From your latest message, it appears that you want the results 4 and 7 not because they are the missing integers in the range of rag.id (1 to 9) but because they are missing from some arbitrary list (2, 3, 4, 5, 7, 8). In that case, if you can put the list into a comma-delimited string, you can do this:
SELECT  TO_NUMBER (column_value)     FROM     XMLTABLE ('2, 3, 4, 5, 7, 8')
MINUS     
SELECT     nid                    FROM     rag
;