Re: a substring question

sanjay rallapally (c696346_at_showme.missouri.edu) wrote:

:       where substr( upper(emp_name),1) LIKE upper(v_address);
        where substr( upper(emp_name),1) LIKE '%' || upper(v_address) || '%';

IAP

--
In an attempt to reduce junk email I use an invalid 'From' address.
My correct email address can can be determined by replacing 'not.valid' with 
'value.net'