Verify if the second character is a letter in SQL

Question

I want to put a condition in my query where I have a column that should contain second position as an alphabet.

How to achieve this?

I've tried with _[A-Z]% in where clause but is not working. I've also tried [A-Z]%.

Any inputs please?

Answer 1

I think you want mysql query. like this

SELECT * FROM table WHERE column REGEXP '^.[A-Za-z]+$'

or sql server

select * from table  where  column like '_[a-zA-Z]%'

Answer 2

You can use regular expression matching in your query. For example:

SELECT * FROM `test` WHERE `name` REGEXP '^.[a-zA-Z].*';

That would match the name column from the test table against a regex that verifies if the second character is either a lowercase or uppercase alphabet letter.

Also see this SQL Fiddle for an example of data it does and doesn't match.

Answer 3

The use of '%[A-Z]%' suggests that you are using SQL Server. If so, you can do this using LIKE:

where col like '_[A-Z]%'

For LIKE patterns, _ represents any character. If the first character needs to be a digit:

where col like '[0-9][A-Z]%'

EDIT:

The above doesn't work in DB2. Instead:

where substr(col, 2, 1) between 'A' and 'Z'

Answer 4

agree with @Gordon Linoff, your ('_[A-Z]%') should work. if not work, kindly add some sample data with your question.

Declare @Table Table
(
    TextCol     Varchar(20)
)

Insert Into @Table(TextCol) Values
    ('23423cvxc43f')
,('2eD97S9')
,('sAgsdsf')
,('3Ss08008')

Select  *
From    @Table As t
Where   t.TextCol Like '_[A-Z]%'