Select with like for multiple results

Question

Hi,
I need to find all the rows of a table that contain the values of another query in the table.
My problem is that the table that I want to verify may contain a list of values and I’m not been able to think on an easy way to do this!
Performance is also an issue because my table2 will have only a short list of values to look for but the table I will search will have a few million rows.
The field that I need to search is a “VARCHAR2(4000 BYTE)” It is not indexed but probably I will need to create an index in it in case the best solution involves it!
My example scenario:
CREATE TABLE TEST_TABLE
(id number,
COL1 CHAR(25));
CREATE TABLE TEST_TABLE2
(COL1 CHAR(25));
insert into TEST_TABLE (id,COL1) values (1,'1;2;3');
insert into TEST_TABLE (id,COL1) values (2,'1');
insert into TEST_TABLE2 (COL1)  values ('1');
insert into TEST_TABLE2 (COL1) values ('2');
select id from TEST_TABLE
where (col1) in ( select col1 from TEST_TABLE2)My results with this are:
Results:
ID
2I need the results to be ( because ID 1 also contains the values from table 2):
Results:
ID
1
2Thanks 
Ricardo Tomás

Answer 1

select  id
  from  test_table t1
  where exists (
                select  1
                  from  test_table2 t2
                  where ';' || trim(t1.col1) || ';' like '%;' || trim(t2.col1) || ';%'
        ID
         1
         2
SQL> Why column COL1 is defined as CHAR? Since CHAR is right blank padded you will have to trim. Use VARCHAR2:
SQL> select  id
  2    from  test_table t1
  3    where exists (
  4                  select  1
  5                    from  test_table2 t2
  6                    where ';' || t1.col1 || ';' like '%;' || t2.col1 || ';%'
  7                 )
  8  /
        ID
         1
         2
SQL> SY.