A situation has arisen where I need to perform a base 36 to base 10 conversion, in the context of a SQL statement. There doesn't appear to be anything built into Oracle 9, or Oracle 10 to address this sort of thing. My Google-Fu, and AskTom suggest creating a pl/sql function to deal with the task. That is not an option for me at this point. I am looking for suggestions on an approach to take that might help me solve this issue.
To put this into a visual form...
WITH
Base36Values AS
(
SELECT '0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ' myBase36 FROM DUAL
),
TestValues AS
(
SELECT '01Z' BASE36_VALUE,
71 BASE10_VALUE FROM DUAL
)
SELECT *
FROM Base36Values,
TestValues
I am looking for something to calculate the value 71, based on the input 01Z.
EDIT - that is backwards... given 01Z translate it to 71.
As a bribe, each useful answer gets a free upvote.
Thanks
Evil.
select sum(position_value) from
(
select power(36,position-1) * case when digit between '0' and '9'
then to_number(digit)
else 10 + ascii(digit) - ascii('A')
end
as position_value
from (
select substr(input_string,length(input_string)+1-level,1) digit,
level position
from (select '01Z' input_string from dual)
connect by level <= length(input_string)
)
)
For T-SQL the following logic will perform the task that the Oracle code above does. This is generic general solution and will support Base-X to Base-10:
select
sum(power(base,pos-1) *
case when substring(cnv,pos,1) between '0' and '9' then
cast(substring(cnv,pos,1) as int)
else 10 + ascii(upper(substring(cnv,pos,1))) - ascii('A') end)
from (values(reverse('01Z'), 36)) as t(cnv,base)
left join (values(1),(2),(3),(4),(5),(6)) as x(pos)
on pos <= len(cnv)
To use with other bases just use:
from (select cnv = reverse('FF'), base=16) as t
or
from (select cnv = reverse('101'), base=2) as t
Note that to support strings longer than 6 you would need to add more values to the position vector.
Related
I need to display oracle as output like this
O
R
A
C
L
E
I tried and the below syntax was my best, but I didn't get the result I want
Select 'oracle' from dual connect by level<=10;
I know we can do this by level clause by I don't know how. And is there any way other than level please share
You want to take a single character from the input string and display it on each line, so use:
WITH cteString AS (SELECT 'oracle' AS TEST_STRING FROM DUAL)
SELECT UPPER(SUBSTR(TEST_STRING, LEVEL, 1))
FROM cteString
CONNECT BY LEVEL <= LENGTH(TEST_STRING);
Best of luck.
You can also solve the problem by using a recursion.
This solution may not necessarily be the best, but should show how to solve the problem otherwise
WITH recursion (word, result)as (
SELECT 'oracle' AS word, substr('oracle',1,1) as result FROM DUAL
union all
select substr(word,2,LENGTH(word)-1), substr(word,2,1)
from recursion
where LENGTH(word) > 1
)
select result from recursion
I need to replace the entire word with 0 if the word has any non-digit character. For example, if digital_word='22B4' then replace with 0, else if digital_word='224' then do not replace.
SELECT replace_funtion(digital_word,'has non numeric character pattern',0,digital_word)
FROM dual;
I tried decode, regexp_instr, regexp_replace but could not come up with the right solution.
Please advise.
Thank you.
the idea is simple - you need check if the value is numeric or not
script:
with nums as
(
select '123' as num from dual union all
select '456' as num from dual union all
select '7A9' as num from dual union all
select '098' as num from dual
)
select n.*
,nvl2(LENGTH(TRIM(TRANSLATE(num, ' +-.0123456789', ' '))),'0',num)
from nums n
result
1 123 123
2 456 456
3 7A9 0
4 098 098
see more articles below to see which way is better to you
How can I determine if a string is numeric in SQL?
https://asktom.oracle.com/pls/asktom/f?p=100:11:0::::P11_QUESTION_ID:15321803936685
How to tell if a value is not numeric in Oracle?
You might try the following:
SELECT CASE WHEN REGEXP_LIKE(digital_word, '\D') THEN '0' ELSE digital_word END
FROM dual;
The regular expression class \D matches any non-digit character. You could also use [^0-9] to the same effect:
SELECT CASE WHEN REGEXP_LIKE(digital_word, '\D') THEN '0' ELSE digital_word END
FROM dual;
Alternately you could see if the value of digital_word is made up of nothing but digits:
SELECT CASE WHEN REGEXP_LIKE(digital_word, '^\d+$') THEN digital_word ELSE '0' END
FROM dual;
Hope this helps.
The fastest way is to replace all digits with null (to simply delete them) and see if anything is left. You don't need regular expressions (slow!) for this, you just need the standard string function TRANSLATE().
Unfortunately, Oracle has to work around their own inconsistent treatment of NULL - sometimes as empty string, sometimes not. In the case of the TRANSLATE() function, you can't simply translate every digit to nothing; you must also translate a non-digit character to itself, so that the third argument is not an empty string (which is treated as a real NULL, as in relational theory). See the Oracle documentation for the TRANSLATE() function. https://docs.oracle.com/cd/E11882_01/server.112/e41084/functions216.htm#SQLRF06145
Then, the result can be obtained with a CASE expression (or various forms of NULL handling functions; I prefer CASE, which is SQL Standard):
with
nums ( num ) as (
select '123' from dual union all
select '-56' from dual union all
select '7A9' from dual union all
select '0.9' from dual
)
-- End of simulated inputs (for testing only, not part of the solution).
-- SQL query begins BELOW THIS LINE. Use your own table and column names.
select num,
case when translate(num, 'z0123456789', 'z') is null
then num
else '0'
end as result
from nums
;
NUM RESULT
--- ------
123 123
-56 0
7A9 0
0.9 0
Note: everything here is in varchar2 data type (or some other kind of string data type). If the results should be converted to number, wrap the entire case expression within TO_NUMBER(). Note also that the strings '-56' and '0.9' are not all-digits (they contain non-digits), so the result is '0' for both. If this is not what you needed, you must correct the problem statement in the original post.
Something like the following update query will help you:
update [table] set [col] = '0'
where REGEXP_LIKE([col], '.*\D.*', 'i')
I have to select only the IDs which have only even digits (an ID looks like: p19 ,p20 etc). That is, p20 is good (both 2 and 0 are even digits); p18 is not.
I thought to use substr to get each number from the IDs and then see if it's even .
select from profs
where to_number(substr(id_prof,2,2))%2=0 and to_number(substr(id_prof,3,2))%2=0;
IF you need all rows consist of 'p' in beginning and even digits on tail It should look like:
select *
from profs
where regexp_like (id_prof, '^p[24680]+$');
with
profs ( prof_id ) as (
select 'p18' from dual union all
select 'p24' from dual union all
select 'p53' from dual
)
-- End of test data; what is above this line is NOT part of the solution.
-- The solution (SQL query) begins here.
select *
from profs
where length(prof_id) = length(translate(prof_id, '013579', '0'));
PROF_ID
-------
p24
This solution should work faster than anything using regular expressions. All it does is to replace 0 with itself and DELETE all odd digits from the input string. (The '0' is included due to a strange but documented behavior of translate() - the third argument can't be empty). If the length of the input string doesn't change after the translation, that means the input string didn't have any odd digits.
where mod(to_number(regexp_replace(id_prof, '[^[:digit:]]', '')),2) = 0
I'm using this code
(SELECT (MAX(CODE) +1 WHERE ISNUMERIC([code]) = 1)
I want to max +1 only my numbers of my column preventing characters characters.
NOTE: THIS QUESTION WAS TAGGED MYSQL WHEN THIS ANSWER WAS POSTED.
You can use substring_index() to split the values and then re-unite them:
(SELECT CONCAT(SUBSTRING_INDEX(MAX(Code), '-', 1), '-',
SUBSTRING_INDEX(MAX(CODE), '-', -1) + 1
)
FROM . . .
WHERE code LIKE '%NEW-1%'
)
This assumes that the wildcards do not have hyphens in them, and that the values after the "1" are all numbers.
Also, this doesn't pad the number is zeroes, but that is a good idea for such codes -- it ensures that they are always the same length and that they sort correctly.
The MAX() function accepts expressions, not just column names:
SELECT MAX(CASE ISNUMERIC(code) WHEN 1 THEN code END)+1 as next_code
FROM (
SELECT '15' AS code
UNION ALL SELECT ' 98 ' AS code
UNION ALL SELECT 'New-45' AS code
) foo
WHERE ISNUMERIC(code)=1;
16
(Link is to SQL Server 2005, docs for SQL Server 2000 are apparently no longer on line, but MAX() belongs to SQL standard anyway.)
Why does
select *
from (
SELECT LEVEL as VAL
FROM DUAL
CONNECT BY LEVEL <= 1000
ORDER BY LEVEL
) n
left outer join (select to_number(trim(alphanumeric_column)) as nr from my_table
where NOT regexp_like (trim(alphanumeric_column),'[^[:digit:]]')) d
on n.VAL = d.nr
where d.nr is null
and n.VAL >= 100
throw a ORA-01722 invalid number (reason is the last row, n.VAL), whereas the similar version with numeric columns im my_table works fine:
select *
from (
SELECT LEVEL as VAL
FROM DUAL
CONNECT BY LEVEL <= 1000
ORDER BY LEVEL
) n
left outer join (select numeric_column as nr from my_table) d
on n.VAL = d.nr
where d.nr is null
and n.VAL >= 100
given that numeric_column is of type number and alphanumeric_column of type nvarchar_2. Note that the upper example works fine without the numerical comparison (n.VAL >= 100).
Does anybody know?
This problem was driving me crazy. I narrowed the problem to a simpler query
SELECT *
FROM (SELECT TO_NUMBER(TRIM (alphanumeric_column)) AS nr
FROM my_table
WHERE NOT REGEXP_LIKE (TRIM (alphanumeric_column), '[^[:digit:]]')) d
WHERE d.nr > 1
With alphanumeric_colum values of ('100','200','XXXX'); Running the above statement gave the "invalid number" error. I then made a slight change to the query to use the CAST function instead of TO_NUMBER:
SELECT *
FROM (SELECT CAST (TRIM (alphanumeric_column) AS NUMBER) AS nr
FROM my_table
WHERE NOT REGEXP_LIKE (TRIM (alphanumeric_column), '[^[:digit:]]')) d
WHERE d.nr > 1
And this correctly returned - 100, 200. I would think that those functions would be similar in behavior. It almost appears as though oracle is trying to evaluate the d.nr > 1 constraint before the view is constructed, which makes no sense. If anyone can shed light on why this is happening, I would be grateful. See SQLFiddle example
UPDATE: I did some more digging, because I don't like not knowing why something just works. I ran EXPLAIN PLAN on both queries and got some interesting results.
For the query that failed, the predicate information looks like this:
1 - filter(TO_NUMBER(TRIM("ALPHANUMERIC_COLUMN"))>1 AND NOT
REGEXP_LIKE (TRIM("ALPHANUMERIC_COLUMN"),'[^[:digit:]]'))
You will notice that the TO_NUMBER function is called first in the AND condition, then
the regexp to exclude alpha values. I am thinking oracle maybe does a short-circuit evaluation with the AND condition, and since it is executing TO_NUMBER first, it fails.
However, when we use the CAST function, the evaluation order is swapped, and the
regexp exclusion is evaluated first. Since for the alpha values, it is false, then the
second part of the AND clause is not evaluated, and the query works.
1 - filter( NOT REGEXP_LIKE (TRIM("ALPHANUMERIC_COLUMN"),'[^[:digit:]
]') AND CAST(TRIM("ALPHANUMERIC_COLUMN") AS NUMBER)>1)
Oracle can be strange sometimes.
I believe when it comes to the Predicate (where) clause, Oracle can/will reorder the entire plan as it sees fit. So with regard to the predicate, it will short-circuit (as OldProgrammer noted) the evaluation however it wants, and you wont be able to guarantee the exact order it occurs.
In your current SQL, you are depending on the predicate to remove non numbers. One option would be to not use "WHERE NOT regexp_like ..." and instead use regexp_substr with coalesce. For example:
create table t_tab2
(
col varchar2(10)
);
create index t_tab2_idx on t_tab2(col);
insert into t_tab2
select level from dual
connect by level <= 100;
insert into t_tab2 values ('123ABC456');
commit;
-- select values > 95 (96->100 exclude non numbers)
select d.* from
(
select COALESCE(TO_NUMBER(REGEXP_SUBSTR(trim(col), '^\d+$')), 0) as nr
from t_tab2
) d
where d.nr > 95;
This should run without throwing invalid number error. Note that the coalesce will return the number 0 for any non numbers coming from the data, you may want to change that based on your needs and data.