How to get partial string in oracle

I need to find all the users in the table that have this pattern

[e-mail]
[e-mail]
[e-mail]

I only know the first 10 digits, the email address can be different.

I did something like this, but I get nothing back

   select * 
   from table where
        substr(user_identifier, 1,10) in (
        '0123456789','0234567980', '0234566666');

Is there another way that I can capture this data in oracle?

Thank you!

Answer

More a comment that a real answer, but with your sample data your code works well:

SQL> with someTable(user_identifier) as
  2  (
  3      select '[e-mail]' from dual union all
  4      select '[e-mail]' from dual union all
  5      select '[e-mail]' from dual union all
  6      select '[e-mail]' from dual
  7  )
  8  select *
  9  from someTable where
 10      substr(user_identifier, 1,10) in ('0123456789','0234567980', '0234566666');

USER_IDENTIFIER
----------------------
[e-mail]
[e-mail]
[e-mail]

3 rows selected.

Say you have a table like this:

create table someTable(user_identifier ) as 
    select '[e-mail]' from dual union all
    select '[e-mail]' from dual union all
    select '[e-mail]' from dual union all
    select '[e-mail]' from dual

SQL> select * from someTable;

USER_IDENTIFIER
-------------------------
[e-mail]
[e-mail]
[e-mail]
[e-mail]

4 rows selected.

Your query would give

USER_IDENTIFIER
-------------------------
[e-mail]
[e-mail]
[e-mail]

3 rows selected.

What do you need instead?