On Tue, 15 Jul 2025, Jeff Ross wrote:
> How about
>
> test:
> select company_name, replace(company_name,'.','') from companies;
>
> update:
> update companies set company_name = replace(company_name,'.','') where
> company_name like '%.';
Jeff,
These contain the table and column names I didn't see in web page examples.
Using update looks better to me.
Many thanks,
Rich