SQL & PL/SQL :: Finding Value Of Bits From Hex Number?
Jan 12, 2012
Version : Oracle Database 11g Enterprise Edition Release 11.2.0.1.0 - 64bit Production
I have a source which is of datatype RAW(1). Basic requirement is to convert it into binary, and then lookup the value of the 4th and 6th bit. Is this possible using simple SQL ?
Eg.
DataBinary
----- -------
EF1110 1111
From this binary output, i need to find the value of 4th and 5th bit.
So if the 4th bit is 1, I need to perform some operations if 0 then some other.
Create table test_hex_to_bin (col1 raw(1));
INSERT into test_hex_to_bin values ('99');
INSERT into test_hex_to_bin values ('EF');
INSERT into test_hex_to_bin values ('AB');
Basically the output should read as
Bit_number Value
----------------
4 1
5 0
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Feb 4, 2012
The code which I am working on consists of an incoming dynamic string which be in the form of binary digits. The max size of the string will be 12 digits. For example, the string can be '111011000001', '000000000000', '111111011111', etc.
I need to find the number of occurences of '111' in the incoming string. Say in the 1st example, result will be 1, in the 2nd example result will be 0, and in the third example, the result will be 3.
I have been trying to capture the string length and replacing the variables '111' to find the number of occurences, but it isn't giving me the result that I want. This is what I have tried
SQL> conn hr/hr
Connected.
SQL> show user
USER is "HR"
[Code]....
I searched the forum and found a similar topic, and following that guideline, I even tried dividing the string with the length of the pattern. It works in some scenarios (the first and second examples mentioned below), while it fails in some scenarios (third example mentioned below)
SQL> select (length('11101110111') - length(replace('11101110111','111','')))/length('111') as occurences from dual;
OCCURENCES
----------
3
SQL> select (length('110111110111') - length(replace('110111110111','111','')))/length('111') as occurences from dual;
OCCURENCES
----------
2
SQL> select (length('111111111111') - length(replace('111111111111','111','')))/length('111') as occurences from dual;
OCCURENCES
----------
SQL>
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and
r1 has values only for c1,c2
r2 has values only for c1,c2,c3,c4
r3 has values only for c1
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Jun 19, 2013
I have the following query where i need to find the max value[code] with t as
( select 'L1' R_nm ,'Data' R_Data , 'Obj' R_Obj , 'Wd' r_prec , '2' val
from dual union all select 'L1' , 'Data', 'Obj' , 'No', '4'
from dualunion all select 'L2' , 'Data', 'Obj' , 'No', '4' from dual )
select t.*, max(val) over(partition by r_nm,r_data,r_obj) rk from t[/code]
My expected output should be
[code]
r_nm r_data r_obj r_prec val rkL1 Data Obj Wd 2 1L1 Data Obj No 4 2L2 Data Obj Yes 1 1
[/code]
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Dec 4, 2011
I'm not really sure why oracle is not finding my Foreing Key, I'm creating an easy set of table for a company and I'm declaring all Primary keys and foreing keys as necessary and this is my
CREATE TABLE Items
(ItemID NUMBER(5) NOT NULL,
Description VARCHAR2(30),
ListPrice NUMBER(9,2) NOT NULL,
QtyOnHand Number(5),
ReorderPoint NUMBER(3),
ReorderQty NUMBER(3),
OnBackOrder Number(3),
CONSTRAINT ItemsPK
[Code] ............
There's three sets of inserts the first two successfully inserts the data NO problem but when i try to insert the data on SupplierItems comes as:
ERROR at line 1:
ORA-02291: integrity constraint (jcarter.SUPPLIERITEMSFK2) violated - parent key not found
I looked on the forum and it appears this happens when someone is doing a Insert ALL but this is not my case..
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Oct 31, 2012
Select * from one:
ID SALARY
----------------------
10 1000
20 2000
30 3200
Select * from two:
ID SALARY
----------------------
10 1000
20 2000
40 4000
10 3000
20 4000
30 3000
When i try to find ID,MAX(SALARY) from these two tables, i am getting this output:
Select id,max(salary)
from
(select * from one
union
select * from two)
group by id
order by max(salary) desc;
OUTPUT:
ID MAX(SALARY)
--------------------------------------
20 4000
40 4000
30 3200
10 3000
I want OUTPUT to be like this:
ID MAX(SALARY)
--------------------------------------
20 4000
40 4000
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sql
UPDATE missions m
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FROM missions m2
WHERE m2.mission_type_id = m.mission_type_id
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From missions m3
ORDER BY m3.mission_id desc)
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example
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CREATE TABLE TEST_DUB ( NAME VARCHAR2(99), V_COUNT NUMBER, FLAG VARCHAR2(1));
Insert into TEST_DUB (NAME, V_COUNT)
Values
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Insert into TEST_DUB
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[Code]....
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[Code]....
It reads:
Suppose that you want to select every employee whose last name has a double vowel(two adjacent occurrences of the same vowel).
Example 4.9 shows how you can do this.
The regular expression ([AEIOU]) represents any vowel. The metacharacter 1 represents the first (and in this case, only) regular expression. The third function parameter, 'i', specifies that the match is case-insensitive.
Example 4.9 Selecting All Employees Whose Last Names Have Double Vowels
SELECT FIRST_NAME, LAST_NAME
FROM EMPLOYEES
WHERE REGEXP_LIKE(LAST_NAME, '([AEIOU])1', 'i');
Result is similar to:
FIRST_NAME LAST_NAME
-------------------- -------------------------
Harrison Bloom
Lex De Haan
Kevin Feeney
Ki Gee
Nancy Greenberg
Danielle Greene
Alexander Khoo
David Lee
8 rows selected.
I don't understand how [AEIOU])1 would find a double vowel(two adjacent occurrences of the same vowel).
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I have contents like below:
BREAD,BISCUIT
BREAD,MILK,BISCUIT
COKE,MILK
MILK,SUGAR
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BREAD,BISCUIT --> [COMBINATION] {BISCUIT} {BREAD} {BISCUIT,BREAD}
BREAD,MILK,BISCUIT --> [COMBINATION] {BISCUIT} {BREAD} {MILK} {BISCUIT,BREAD} {BISCUIT,MILK} {BREAD,MILK} {BISCUIT,BREAD,MILK}
COKE,MILK --> [COMBINATION] {COKE} {MILK} {COKE,MILK}
MILK,SUGAR --> [COMBINATION] {MILK} {SUGAR} {MILK,SUGAR}
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[Code]....
writing query to find unique combination like above? What I have tried is as below:
create table test (row_no number, col_no number, item varchar2(50))
/
insert into test values (1,1,'BREAD');
insert into test values (1,2,'BISCUIT');
insert into test values (2,1,'BREAD');
[Code]....
But I am not able to form the exact query. let me know if this can't be done through a single query.
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-------------------
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---------------------
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