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


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