Display all the names of the employees whose names having letter A as 2nd occurrence without using like operator - sql

Question

For the above query, I am using regexp_count but in SQL command line I am getting regexp_count :invaild identifier;
select * from table_name WHERE (REGEXP_COUNT(column_name, 'A')) >2;
Is this query works?

Answer 1

I'm not sure what you really want; is it to return names whose 2nd letter is "a", or names that have two or more letters "a" within.
Anyway, pick the one you find appropriate.
SQL> create table test as
2 (select 'saritha' col from dual union all
3 select 'mamatha' from dual union all
4 select 'vaisnavi' from dual union all
5 select 'sai' from dual union all
6 select 'vijaya' from dual union all
7 select 'kumar' from dual
8 );
Table created.
2 or more letters "a":
SQL> select col
2 from test
3 where regexp_count(col, 'a') >= 2;
COL
--------
saritha
mamatha
vaisnavi
vijaya
2nd letter is "a":
SQL> select col
2 from test
3 where substr(col, 2, 1) = 'a';
COL
--------
saritha
mamatha
vaisnavi
sai
SQL>
On Oracle 10g, which doesn't support REGEXP_COUNT function, one option is to replace all letters a with an empty string (basically, you'd remove all letters a) and fetch rows whose difference of full column length and "replaced" column length is >= 2. Something like this:
SQL> select col
2 from test
3 where length(col) - length(replace(col, 'a', '')) >= 2;
COL
--------
saritha
mamatha
vaisnavi
vijaya
SQL>