How to find a rows in column with two letters in oracle

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i have a table like,
ID NAME
1 AB JOHN
2 MICHAEL JOHNSON
3 ST SACHIN

I want to generate like(I want query in oracle)
ID NAME
1 AB JOHN
3 ST SACHIN

Plz help me from this

What I have tried:

SELECT *
FROM Test_V
WHERE regexp_like (name , '[]') ;

Posted 4-Sep-17 6:17am

Member 11416690

Updated 4-Sep-17 23:44pm

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itsmypassion · Answer 1 · 2017-09-04T21:01:00

Solution 1

SQL

Select * from Test_V Where Locate(' ', Name)=3

Posted 4-Sep-17 21:01pm

itsmypassion

RAMASWAMY EKAMBARAM · Answer 2 · 2017-09-04T23:44:00

Member 11416690 is on the right track using regexp_like()

Try this:
select *
from test_v
where regexp_like(' ' || name || ' ', [^[:alpha:]]+[[:alpha:]]{2}[^[:alpha:]]+)

The above will test for 2 consecutive alpha characters (letters) preceded and succeeded by at least 1 non alphabetic character. As I have not done any real testing of the query in Oracle, I am not sure whether Oracle will return a match if the 2 characters are at the start/ end of the string and that is why the prefix and suffix spaces are added to 'name' (column value). The above should match even if the 2 consecutive letters occur in the middle of the string. If you are looking to match uppercase characters use [[:upper:]]{2} but do not alter [^[:alpha:]]+.
locate(' ', name) = 3 will also match A. JOHN and S. SACHIN.