Anyway, got asked if there was a way to return a resultset of data based on the leading character, the use case was to ignore all strings starting with a,b, or d and return e to z.
Fairly straight forward but a good SQL to have.
I simply grab the first character in the varchar and compare it's ASCII value to the ASCII value D.
SQL> select * from t;
A
----------
Atest
Btest
Etest
Htest
Wtest
Dtest
dtest
SQL> with data as (
2 select ascii(upper(substr(a,1,1))) a_val,a from t
3 )
4 select * from data where a_val not between ascii('A') and ascii('D') order by a
5 /
A_VAL A
---------- ----------
69 Etest
72 Htest
87 Wtest
SQL>
4 comments:
I think your result is not correct, I got a different result:
Connected to:
Oracle Database 10g Enterprise Edition Release 10.2.0.1.0 - Production
With the Partitioning, OLAP and Data Mining options
SQL> with t as (
2 select 'Atest' a from dual union all
3 select 'Btest' from dual union all
4 select 'Etest' from dual union all
5 select 'Htest' from dual union all
6 select 'Wtest' from dual union all
7 select 'Dtest' from dual union all
8 select 'dtest' from dual),
9 data as (
10 select ascii (upper (substr (a, 1, 1) ) ) a_val, a
11 from t)
12 select ascii ('A'), ascii ('D'), data.*
13 from data
14 where a_val not between ascii ('A') and ascii ('D')
15 order by a;
ASCII('A') ASCII('D') A_VAL A
---------- ---------- ---------- -----
65 68 69 Etest
65 68 72 Htest
65 68 87 Wtest
I think the values of Dtest and dtest should not be displayed because their ascii value (68) is between 65 and 68.
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