Is there a way to display a sql LIKE-operator in relational algebra?
right now I have done it like this:
σfirstname LIKE ('B%')person
But I'm not sure if this is the right way to go.
Thanks in advance for the help.
Here is a similar question covered earlier on this site regarding how to use LIKE in relational algebra. In that case, the professor had indicated that it was ok to use LIKE just as in SQL.