Simple question about regular expression

Question

Hi
I have a little problem with
select regexp_substr('123 Mapla Avenue','[a-z]') my_test from dual;
answer: M
I excecute this query in SQLPlus and SQL Developer result is this same.
select regexp_substr('123 Mapla Avenue','[M]') my_test from dual;
answer: M
select regexp_substr('123 Mapla Avenue','[a]') my_test from dual;
answer: a
I used oracle 10g
Thanks for your help

Answer 1

hm wrote:
In the oracle documentation of regexp_substr you can find:Do not confuse pattern and sort. Pattern [a-z] means any lowercase letter. REGEXP_SUBSTR parameter match_param value i tells REGEXP to treat uppercase letters same as lowercase letters and vice versa. And setting NLS_SORT can do the same. As you can see it is not that straight-forward. To make it transparent use exact pattern you need. In this particular case use:
select regexp_substr('123 Mapla Avenue','[[:alpha:]]') my_test from dual;where class [:alpha:] is POSIX predefined class of all letters (regardless of case). This way you are not dependent of client side settings like NLS_SORT and the above will always return first letter within a string. If you want first uppercase letter use:
select regexp_substr('123 Mapla Avenue','[[:upper:]]') my_test from dual;Or, for first lowercase letter:
SQL> alter session set nls_sort=binary;
Session altered.
SQL> select regexp_substr('123 Mapla Avenue','[a-z]') my_test from dual;
M
a
SQL> select regexp_substr('123 Mapla Avenue','[[:lower:]]') my_test from dual;
M
a
SQL> alter session set nls_sort=binary_ci;
Session altered.
SQL> select regexp_substr('123 Mapla Avenue','[a-z]') my_test from dual;
M
M
SQL> select regexp_substr('123 Mapla Avenue','[[:lower:]]') my_test from dual;
M
a
SQL> SY.