Greetings,

I am currently using Oracle9i Enterprise Edition Release 9.2.0.4.0. I
have a table with following data

Table 1 (Sample data)
a12345
A123423
g13452
G452323
h34423
r34323
b23232
n232323

I am currently using this as a subquery in one of the query. As per a
new request I have to now exclude all values which start with h, b or
n followed by numeric values.

So end result the subquery should give me is

Table 1 (Sample data)
a12345
A123423
g13452
G452323
r34323

I am little stumped on this for now. Could not get it right in my
query. Can anyone please advise here. Let me know if any more
information is needed from my side.

Note: The starting character in all values can sometimes in "lower
case" or sometimes in "upper case".

TIA

On Dec 29, 8:54*am, Pankaj <harpreet.n... (AT) gmail (DOT) com> wrote:
Some people prefer that you post create table, insert data and what
sql statements you've tried, so they can quickly ramp up a test
environment. Is this school or work? (If school, people are usually
willing to give you hints on how to figure it out, not do it for you).

You could substr, uppercase and notinlist for the first character in
your where statement, and there are several ways to check the rest for
numeric, like http://www.adp-gmbh.ch/ora/plsql/hel...s_numeric.html

jg
--
@home.com is bogus.
Two weeks... http://it.slashdot.org/story/09/12/2...2010?art_pos=1

On Dec 29, 6:30*pm, joel garry <joel-ga... (AT) home (DOT) com> wrote:
I've used TRANSLATE() a lot in tasks similar to the one described by
op.

You may want to take a look at it.

HTH.

Cheers.

Carlos.

On Dec 29, 11:54*am, Pankaj <harpreet.n... (AT) gmail (DOT) com> wrote:
I agree with Joel's comments. Let's see if there is a hard way to do
this.

CREATE TABLE T10(HOMEWORK VARCHAR2(20));

INSERT INTO T10 VALUES ('a12345');
INSERT INTO T10 VALUES ('A123423');
INSERT INTO T10 VALUES ('g13452');
INSERT INTO T10 VALUES ('G452323');
INSERT INTO T10 VALUES ('h34423');
INSERT INTO T10 VALUES ('r34323');
INSERT INTO T10 VALUES ('b23232');
INSERT INTO T10 VALUES ('n232323');
INSERT INTO T10 VALUES ('NB151517');
INSERT INTO T10 VALUES ('C0151517');
INSERT INTO T10 VALUES ('f9151517');
INSERT INTO T10 VALUES ('HE4423');

COMMIT;

Note that I added a couple of extra rows just for fun.

Let's look at the ASCII values of the first and second characters:
SELECT
HOMEWORK,
ASCII(SUBSTR(HOMEWORK,1,1)) ASC_VAL1,
ASCII(SUBSTR(HOMEWORK,2,1)) ASC_VAL2
FROM
T10;

HOMEWORK ASC_VAL1 ASC_VAL2
---------- ---------- ----------
a12345 97 49
A123423 65 49
g13452 103 49
G452323 71 52
h34423 104 51
r34323 114 51
b23232 98 50
n232323 110 50
NB151517 78 66
C0151517 67 48
f9151517 102 57
HE4423 72 69

OK, I see the ones that we want to exclude, let's build a matrix:
SELECT
HOMEWORK,
ASCII(SUBSTR(HOMEWORK,1,1)) ASC_VAL1,
ASCII(SUBSTR(HOMEWORK,2,1)) ASC_VAL2,
DECODE(ASCII(SUBSTR(HOMEWORK,1,1)),
104,1,72,1,66,1,98,1,78,1,110,1,0) IS_EXC1,
DECODE(SIGN(ASCII(SUBSTR(HOMEWORK,2,1))-47),1,DECODE(SIGN(ASCII
(SUBSTR(HOMEWORK,2,1))-58),-1,1,0),0) IS_EXC2
FROM
T10;

HOMEWORK ASC_VAL1 ASC_VAL2 IS_EXC1 IS_EXC2
---------- ---------- ---------- ---------- ----------
a12345 97 49 0 1
A123423 65 49 0 1
g13452 103 49 0 1
G452323 71 52 0 1
h34423 104 51 1 1
r34323 114 51 0 1
b23232 98 50 1 1
n232323 110 50 1 1
NB151517 78 66 1 0
C0151517 67 48 0 1
f9151517 102 57 0 1
HE4423 72 69 1 0

If there is a 1 in both of the right-most columns, then the row should
be eliminated. What is the easiest way to tell if there is a 1 in
both columns? Multiply the column values together, and if we receive
a product of 1 then the row should be excluded:
SELECT
*
FROM
(SELECT
HOMEWORK,
ASCII(SUBSTR(HOMEWORK,1,1)) ASC_VAL1,
ASCII(SUBSTR(HOMEWORK,2,1)) ASC_VAL2,
DECODE(ASCII(SUBSTR(HOMEWORK,1,1)),
104,1,72,1,66,1,98,1,78,1,110,1,0) IS_EXC1,
DECODE(SIGN(ASCII(SUBSTR(HOMEWORK,2,1))-47),1,DECODE(SIGN(ASCII
(SUBSTR(HOMEWORK,2,1))-58),-1,1,0),0) IS_EXC2
FROM
T10)
WHERE
IS_EXC1*IS_EXC2<>1;

HOMEWORK ASC_VAL1 ASC_VAL2 IS_EXC1 IS_EXC2
---------- ---------- ---------- ---------- ----------
a12345 97 49 0 1
A123423 65 49 0 1
g13452 103 49 0 1
G452323 71 52 0 1
r34323 114 51 0 1
NB151517 78 66 1 0
C0151517 67 48 0 1
f9151517 102 57 0 1
HE4423 72 69 1 0


Something tells me you want to do it the easy way. See if you can do
anything with these functions:
REGEXP_INSTR
http://download.oracle.com/docs/cd/B...ctions129..htm

REGEXP_SUBSTR
http://download.oracle.com/docs/cd/B...ctions131..htm

*Always* post the DDL and DML to re-create your problem, and show us
what you have tried previously.

Charles Hooper
Co-author of "Expert Oracle Practices: Oracle Database Administration
from the Oak Table"
http://hoopercharles.wordpress.com/
IT Manager/Oracle DBA
K&M Machine-Fabricating, Inc.

On Dec 30, 5:31*am, Charles Hooper <hooperc2... (AT) yahoo (DOT) com> wrote:
LOL, you should write a book! "Bad SQL! Bad, bad!"

Watch those versions :-)

(And thanks Carlos, I should've thought of that first. TIMTOWTDI)

jg
--
@home.com is bogus.
http://arstechnica.com/web/news/2009...rmance-art.ars

On Dec 30, 12:28*pm, joel garry <joel-ga... (AT) home (DOT) com> wrote:
That would be an interesting title for a book. Take a somewhat simple
request and see how many different (or overly complex) solutions may
be generated for the request.

More specifically on your second point, regular expressions are not
available in Oracle 9i R2 - for some reason I thought that they were
introduced with Oracle 9i R1 (I even performed a search to verify - I
should have clicked one of the links). After seeing your post, I
searched again and found a couple interesting articles for those
people running Oracle 10g R1 and above:
http://download.oracle.com/owsf_2003...Gennick_04.ppt
http://download.oracle.com/owsf_2003/40105.doc

Charles Hooper
Co-author of "Expert Oracle Practices: Oracle Database Administration
from the Oak Table"
http://hoopercharles.wordpress.com/
IT Manager/Oracle DBA
K&M Machine-Fabricating, Inc.

On Dec 30, 4:49*pm, Charles Hooper <hooperc2... (AT) yahoo (DOT) com> wrote:
Thanks Everyone.

Carlos/Joe: I tried TRANSLATE option and it works.
Charles: I will go ahead with your option for now. Can you please
detail me on what the below expression is doing.

DECODE(SIGN(ASCII(SUBSTR(HOMEWORK,2,1))-47),1,DECODE(SIGN(ASCII
(SUBSTR(HOMEWORK,2,1))-58),-1,1,0),0) IS_EXC2

TIA.

On 31.12.2009 19:58, Pankaj wrote:
It checks, whether the second character in the column HOMEWORK
represents a digit. You can look at the results of the query
with t as (
select chr(32)||chr(rownum + 31) c from dual
connect by level <= 128-32
)
select c,
decode(sign(ascii(substr(c,2,1))-47),1,decode(sign(ascii
(substr(c,2,1))-58),-1,1,0),0) is_exc2
from t


Just to mention another approach regarding your question:

SQL> with t as (
2 select 'a12345' c from dual union all
3 select 'A123423' from dual union all
4 select 'g13452' from dual union all
5 select 'G452323' from dual union all
6 select 'h34423' from dual union all
7 select 'r34323' from dual union all
8 select 'b23232' from dual union all
9 select 'n' from dual union all
10 select 'n232323' from dual
11 )
12 -- End test data
13 select c
14 from t
15 where not lower(rtrim(c,'0123456789')) in ('h','b','n')
16 /

C
-------
a12345
A123423
g13452
G452323
r34323


Best regards

Maxim

On Dec 31, 1:58*pm, Pankaj <harpreet.n... (AT) gmail (DOT) com> wrote:
The numbers 0 through 9 have ASCII values ranging from 48 to 57.
* Obtain the second character in the column: SUBSTR(HOMEWORK,2,1)
* Use the ASCII function to find the ASCII value of the second
character
* Subtract 47 from the ASCII value for the second character
* If the difference is greater than 0, then:
** Subtract 58 from that ASCII value
** If the difference is less than 0, then we found an ASCII value
between 48 and 57 - therefore the second character must be a number
*** Return the number 1 if the ASCII value is between 48 and 57,
otherwise return 0

A CASE structure could be used rather than the cumbersome nested
DECODE and SIGN statements. A CASE structure will be easier to
maintain:
SELECT
CASE WHEN ASCII(SUBSTR(HOMEWORK,2,1)) >= 48
AND ASCII(SUBSTR(HOMEWORK,2,1)) <= 57 THEN 1
ELSE 0 END IS_EXC2
FROM
T10;

You could transform this section to a CASE structure also:
DECODE(ASCII(SUBSTR(HOMEWORK,1,1)),104,1,72,1,66,1 ,98,1,78,1,110,1,0)
IS_EXC1

SELECT
CASE ASCII(SUBSTR(HOMEWORK,1,1))
WHEN 104 THEN 1
WHEN 72 THEN 1
WHEN 66 THEN 1
WHEN 98 THEN 1
WHEN 78 THEN 1
WHEN 110 THEN 1
ELSE 0 END IS_EXC1
FROM
T10;

Finally, you could combine the two CASE structures in the WHERE
clause:
SELECT
HOMEWORK,
ASCII(SUBSTR(HOMEWORK,1,1)) ASC_VAL1,
ASCII(SUBSTR(HOMEWORK,2,1)) ASC_VAL2
FROM
T10
WHERE
(CASE ASCII(SUBSTR(HOMEWORK,1,1))
WHEN 104 THEN 1
WHEN 72 THEN 1
WHEN 66 THEN 1
WHEN 98 THEN 1
WHEN 78 THEN 1
WHEN 110 THEN 1
ELSE 0 END) *
(CASE WHEN ASCII(SUBSTR(HOMEWORK,2,1)) >= 48
AND ASCII(SUBSTR(HOMEWORK,2,1)) <= 57 THEN 1
ELSE 0 END) = 0;

HOMEWORK ASC_VAL1 ASC_VAL2
---------- ---------- ----------
a12345 97 49
A123423 65 49
g13452 103 49
G452323 71 52
r34323 114 51
NB151517 78 66
C0151517 67 48
f9151517 102 57
HE4423 72 69

There are probably several other ways to solve this problem.

Charles Hooper
Co-author of "Expert Oracle Practices: Oracle Database Administration
from the Oak Table"
http://hoopercharles.wordpress.com/
IT Manager/Oracle DBA
K&M Machine-Fabricating, Inc.

On Dec 31, 3:30*pm, Maxim Demenko <mdeme... (AT) gmail (DOT) com> wrote:
Nice solution! I did not even think of using RTRIM to strip off the
characters at the right of the string when those characters are found
in the string. You did not even need to divide by 0 to produce the
desired result. :-)

Charles Hooper
Co-author of "Expert Oracle Practices: Oracle Database Administration
from the Oak Table"
http://hoopercharles.wordpress.com/
IT Manager/Oracle DBA
K&M Machine-Fabricating, Inc.