RSA Admin

Using a Parameter in a LIKE Statement

Discussion created by RSA Admin Employee on Oct 21, 2011
Latest reply on Oct 25, 2011 by RSA Admin
Is there any way to use a parameter within a LIKE statement. For example: WHERE Message LIKE '%$(parameter)%' When I try this, I receive an error message because the $(parameter) is expanded to '$(parameter)', and the single quotes screws up the syntax. So, for example, if the parameter equals test, you end up getting: WHERE Message LIKE '%'test'%' which throws a syntax error. Thanks, Brian

Outcomes