I have a table with a
I see two ways doing this:
If you allow only a few extra characters than you can prepare a string which is stripped from these extra characters and you use the LIKE operator you normally would
select * from phoneTable where replace(replace(phone, '+', ''), '-', '') LIKE '%123%'
Of course you need as many replace calls as the number of allowed extra characters
You use regular expressions, let's say you are searching for pattern 123
select * from phoneTable where phone REGEXP '.*1[^0-9]*2[^0-9]*3'