How to get the rows from a table having some column has any letter

Hi All,
suppose i have a table having columns id(number), code(varchar).
code has alphanumeric characters (ex. ABC123, 67B56 etc).
some codes are only numbers (2344, 7898 etc).
how can i get the rows which have alphabets in the code.
ex:
id code
1 AB45
2 456
3 890
4 67B7
how can i write a query such that it should give me the ids 1 and 4 (as they have alphabets in code)
thanks in advance to all

Thanks to one and all.
i am gettig my required output.
But i have a doubt in the operator.
If i add or remove '[]' in the operator, i am getting different ouputs.
There is a count difference in the result of the operators used.
REGEXP_LIKE(<column>,'[[:lower:]]')
REGEXP_LIKE(<column>,'[[[:lower:]]]')
REGEXP_LIKE(<column>,'[:lower:]')
Can anybody please explain what is the difference in using '[]', in the operator?
What is the correct syntax, whether i have to use two '[]'s or one '[]'.
Also, can i use REGEXP_LIKE() in oracle 8i version.( I am unable to use the operator in 8i)?
Any query to get the required output in 8i version?
Thanks in advance to all.

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