I have below requirement.
Input is like as below.
Create table Numbers
(
Num int
)
Insert into Numbers
values (1),(2),(3),(4),(5),(6),(7),(8),(9),(10),(11),(12),(13),(14),(15)
Create table FromTo
(
FromNum int
,ToNum int
)
Select * From FromTo
Output should be as below.
FromNum ToNum
1 5
6 10
11 15
Actual Requirement is as below.
I need to load the data for a column into a table which will have thousands of records with different no's.
Consider like below.
1,2,5,7,9,11,15,34,56,78,98,123,453,765 etc..
I need to load these into other table which is having FROM and TO columns with the intervals of 5000. For example in the first 5000 if i have the no's till 3000, my 1st row should have FromNo as 1 and ToNum as 3000. second row: if the data is not having till 10000 and the next no started as 12312(This is the 2nd Row FromNum) the ToNum value should be +5000 i.e 17312. Here also if we don't have the no's data till 17312 it need to consider the ToNum between the 12312 and 17312
Output should be as below.
FromNum ToNum
1 3205
1095806 1100805
1100808 1105806
1105822 1110820
Can you guys please help me with the solution for the above.
Thanks in advance.
What you may try in this situation is to group data and get the expected results:
DECLARE #interval int = 5
INSERT INTO FromTo (FromNum, ToNum)
SELECT MIN(Num) AS FromNum, MAX(Num) AS ToNum
FROM Numbers
GROUP BY (Num - 1) / #interval
The below TSQL statements are returning different values based on the order of #Size and #Value.The first statement returns 1687.500000 but the second one 1687.600000.
I am guessing it's because of some rounding but I can't really figure out myself. Any help would be really appreciated.
DECLARE #Amount DECIMAL(20,4) = 2,
#PriceDiff DECIMAL(25,10) = 0.421875,
#Size DECIMAL(16,4)= 200000.0000,
#Value DECIMAL(25,15)= 0.010000000000000
SELECT #Amount * #PriceDiff * #Size * #Value AS FinalValue
SELECT #Amount * #PriceDiff * #Value * #Size AS FinalValue
From Management studio
The reason you are experiencing the rounding error is due to the way that SQL Server determines the result of multiplication on a Decimal type's precicion and scale. see here
Also, SQL Server's order of operation for the same/equal-precedence operators is LTR .
Given that the first step is to multiple #Amount * #PriceDiff. According to that link the precision and scale would be:
precision = 20 + 25 + 1 = 46
scale = 4 + 10 = 14
resulting data type = Decimal(46, 14)
This result is over the max allowed precision for a Decimal so things get a little sticky. At the bottom of that link you will see:
The result precision and scale have an absolute maximum of 38. When a result precision is greater than 38, the corresponding scale is reduced to prevent the integral part of a result from being truncated.
Reading up on this further you'll find that instead of just lopping off all of the decimal places to make a Decimal(38,0) or allowing all precision to decimal with a Decimal(38,38) SQL server makes a big fat guess and makes it a Decimal(38,6).
All is well and good though as our result so far is 0.84375 and that fits fine into our new Decimal(38,6) container.
You can see now that if we multiply this by your #Size we will still be within the limits of a Decimal(38,6) with a result of 168750. So we are still good even with the Precision and Scale math and the resulting rounding to a scale of 6 that will occur.
However, if we take that 0.84375 result and multiple by it #Value we get:
precision = 38 + 25 + 1 = 64
scale = 6 + 15 = 21
result = Decimal(64, 21)
Which means we are back at forcing it into a Decimal(38,6)... And 0.0084375 doesn't fit, so it's rounded to 0.08438.
Using SQL Server 2008.
Let's say I have 3 numbers to store.
First number is any 6bit number
Second is any 4bit
Third is also any 6bit number
For example:
declare #firstNumber_6bit tinyint = 50
declare #secondNumber_4bit tinyint = 7
declare #thirdNumber_6bit tinyint = 63
I need to store those 3 numbers in a 2 byte binary (16bit) variable. So, the binary values for the example are:
50 is 110010,
7 is 0111
63 is 111111
The 2 bytes to store are 11001001 11111111
So either one of the following lines should store those values:
declare #my3NumbersIn2Bytes binary(2) = 51711
or
declare #my3NumbersIn2Bytes binary(2) = 0xC9FF
(sorry if I'm messing the byte order in big / little endian, but that's not the point).
Storing and retriving those numbers is a trivial task with .net CLR, but I'm trying to solve this in pure T-SQL and as we know there is no bit shifting in SQL Server. I saw a lot of examples out there that use memory tables to solve problems like these, but that seems totally overkill for doing a simple bit shift... I was thinking something more like a substring for bits could do the trick. I just want to be sure there's no other way to solve this before going the overkill way.
So my question is, what is the most effective way to store those 3 numbers and recover them?
Thanks.-
declare #firstNumber_6bit tinyint = 50
declare #secondNumber_4bit tinyint = 7
declare #thirdNumber_6bit tinyint = 63
declare #my3NumbersIn2Bytes binary(2)
select #my3NumbersIn2Bytes = #firstNumber_6bit*1024 + #secondNumber_4bit *64 + #thirdNumber_6bit
--extract values
select #firstNumber_6bit = #my3NumbersIn2Bytes/1024
select #secondNumber_4bit = (#my3NumbersIn2Bytes%1024)/64
select #thirdNumber_6bit = (#my3NumbersIn2Bytes%1024)%64
select convert(varchar(max), #my3NumbersIn2Bytes, 1) -- just for display
, #firstNumber_6bit
, #secondNumber_4bit
, #thirdNumber_6bit
SQL Fiddle Demo
I have to count the digits after the decimal point in a database hosted by a MS Sql Server (2005 or 2008 does not matter), in order to correct some errors made by users.
I have the same problem on an Oracle database, but there things are less complicated.
Bottom line is on Oracle the select is:
select length( substr(to_char(MY_FIELD), instr(to_char(MY_FILED),'.',1,1)+1, length(to_char(MY_FILED)))) as digits_length
from MY_TABLE
where the filed My_filed is float(38).
On Ms Sql server I try to use:
select LEN(SUBSTRING(CAST(MY_FIELD AS VARCHAR), CHARINDEX('.',CAST(MY_FILED AS VARCHAR),1)+1, LEN(CAST(MY_FIELD AS VARCHAR)))) as digits_length
from MY_TABLE
The problem is that on MS Sql Server, when i cast MY_FIELD as varchar the float number is truncated by only 2 decimals and the count of the digits is wrong.
Can someone give me any hints?
Best regards.
SELECT
LEN(CAST(REVERSE(SUBSTRING(STR(MY_FIELD, 13, 11), CHARINDEX('.', STR(MY_FIELD, 13, 11)) + 1, 20)) AS decimal))
from TABLE
I have received from my friend a very simple solution which is just great. So I will post the workaround in order to help others in the same position as me.
First, make function:
create FUNCTION dbo.countDigits(#A float) RETURNS tinyint AS
BEGIN
declare #R tinyint
IF #A IS NULL
RETURN NULL
set #R = 0
while #A - str(#A, 18 + #R, #r) <> 0
begin
SET #R = #R + 1
end
RETURN #R
END
GO
Second:
select MY_FIELD,
dbo.countDigits(MY_FIELD)
from MY_TABLE
Using the function will get you the exact number of digits after the decimal point.
The first thing is to switch to using CONVERT rather than CAST. The difference is, with CONVERT, you can specify a format code. CAST uses whatever the default format code is:
When expression is float or real, style can be one of the values shown in the following table. Other values are processed as 0.
None of the formats are particularly appealing, but I think the best for you to use would be 2. So it would be:
CONVERT(varchar(25),MY_FIELD,2)
This will, unfortunately, give you the value in scientific notation and always with 16 digits e.g. 1.234567890123456e+000. To get the number of "real" digits, you need to split this number apart, work out the number of digits in the decimal portion, and offset it by the number provided in the exponent.
And, of course, insert usual caveats/warnings about trying to talk about digits when dealing with a number which has a defined binary representation. The number of "digits" of a particular float may vary depending on how it was calculated.
I'm not sure about speed. etc or the elegance of this code. it was for some ad-hoc testing to find the first decimal value . but this code could be changed to loop through all the decimals and find the last time a value was greater than zero easily.
DECLARE #NoOfDecimals int = 0
Declare #ROUNDINGPRECISION numeric(32,16) = -.00001000
select #ROUNDINGPRECISION = ABS(#ROUNDINGPRECISION)
select #ROUNDINGPRECISION = #ROUNDINGPRECISION - floor(#ROUNDINGPRECISION)
while #ROUNDINGPRECISION < 1
Begin
select #NoOfDecimals = #NoOfDecimals +1
select #ROUNDINGPRECISION = #ROUNDINGPRECISION * 10
end;
select #NoOfDecimals
I need a different random number for each row in my table. The following seemingly obvious code uses the same random value for each row.
SELECT table_name, RAND() magic_number
FROM information_schema.tables
I'd like to get an INT or a FLOAT out of this. The rest of the story is I'm going to use this random number to create a random date offset from a known date, e.g. 1-14 days offset from a start date.
This is for Microsoft SQL Server 2000.
Take a look at SQL Server - Set based random numbers which has a very detailed explanation.
To summarize, the following code generates a random number between 0 and 13 inclusive with a uniform distribution:
ABS(CHECKSUM(NewId())) % 14
To change your range, just change the number at the end of the expression. Be extra careful if you need a range that includes both positive and negative numbers. If you do it wrong, it's possible to double-count the number 0.
A small warning for the math nuts in the room: there is a very slight bias in this code. CHECKSUM() results in numbers that are uniform across the entire range of the sql Int datatype, or at least as near so as my (the editor) testing can show. However, there will be some bias when CHECKSUM() produces a number at the very top end of that range. Any time you get a number between the maximum possible integer and the last exact multiple of the size of your desired range (14 in this case) before that maximum integer, those results are favored over the remaining portion of your range that cannot be produced from that last multiple of 14.
As an example, imagine the entire range of the Int type is only 19. 19 is the largest possible integer you can hold. When CHECKSUM() results in 14-19, these correspond to results 0-5. Those numbers would be heavily favored over 6-13, because CHECKSUM() is twice as likely to generate them. It's easier to demonstrate this visually. Below is the entire possible set of results for our imaginary integer range:
Checksum Integer: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
Range Result: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 0 1 2 3 4 5
You can see here that there are more chances to produce some numbers than others: bias. Thankfully, the actual range of the Int type is much larger... so much so that in most cases the bias is nearly undetectable. However, it is something to be aware of if you ever find yourself doing this for serious security code.
When called multiple times in a single batch, rand() returns the same number.
I'd suggest using convert(varbinary,newid()) as the seed argument:
SELECT table_name, 1.0 + floor(14 * RAND(convert(varbinary, newid()))) magic_number
FROM information_schema.tables
newid() is guaranteed to return a different value each time it's called, even within the same batch, so using it as a seed will prompt rand() to give a different value each time.
Edited to get a random whole number from 1 to 14.
RAND(CHECKSUM(NEWID()))
The above will generate a (pseudo-) random number between 0 and 1, exclusive. If used in a select, because the seed value changes for each row, it will generate a new random number for each row (it is not guaranteed to generate a unique number per row however).
Example when combined with an upper limit of 10 (produces numbers 1 - 10):
CAST(RAND(CHECKSUM(NEWID())) * 10 as INT) + 1
Transact-SQL Documentation:
CAST(): https://learn.microsoft.com/en-us/sql/t-sql/functions/cast-and-convert-transact-sql
RAND(): http://msdn.microsoft.com/en-us/library/ms177610.aspx
CHECKSUM(): http://msdn.microsoft.com/en-us/library/ms189788.aspx
NEWID(): https://learn.microsoft.com/en-us/sql/t-sql/functions/newid-transact-sql
Random number generation between 1000 and 9999 inclusive:
FLOOR(RAND(CHECKSUM(NEWID()))*(9999-1000+1)+1000)
"+1" - to include upper bound values(9999 for previous example)
Answering the old question, but this answer has not been provided previously, and hopefully this will be useful for someone finding this results through a search engine.
With SQL Server 2008, a new function has been introduced, CRYPT_GEN_RANDOM(8), which uses CryptoAPI to produce a cryptographically strong random number, returned as VARBINARY(8000). Here's the documentation page: https://learn.microsoft.com/en-us/sql/t-sql/functions/crypt-gen-random-transact-sql
So to get a random number, you can simply call the function and cast it to the necessary type:
select CAST(CRYPT_GEN_RANDOM(8) AS bigint)
or to get a float between -1 and +1, you could do something like this:
select CAST(CRYPT_GEN_RANDOM(8) AS bigint) % 1000000000 / 1000000000.0
The Rand() function will generate the same random number, if used in a table SELECT query. Same applies if you use a seed to the Rand function. An alternative way to do it, is using this:
SELECT ABS(CAST(CAST(NEWID() AS VARBINARY) AS INT)) AS [RandomNumber]
Got the information from here, which explains the problem very well.
Do you have an integer value in each row that you could pass as a seed to the RAND function?
To get an integer between 1 and 14 I believe this would work:
FLOOR( RAND(<yourseed>) * 14) + 1
If you need to preserve your seed so that it generates the "same" random data every time, you can do the following:
1. Create a view that returns select rand()
if object_id('cr_sample_randView') is not null
begin
drop view cr_sample_randView
end
go
create view cr_sample_randView
as
select rand() as random_number
go
2. Create a UDF that selects the value from the view.
if object_id('cr_sample_fnPerRowRand') is not null
begin
drop function cr_sample_fnPerRowRand
end
go
create function cr_sample_fnPerRowRand()
returns float
as
begin
declare #returnValue float
select #returnValue = random_number from cr_sample_randView
return #returnValue
end
go
3. Before selecting your data, seed the rand() function, and then use the UDF in your select statement.
select rand(200); -- see the rand() function
with cte(id) as
(select row_number() over(order by object_id) from sys.all_objects)
select
id,
dbo.cr_sample_fnPerRowRand()
from cte
where id <= 1000 -- limit the results to 1000 random numbers
select round(rand(checksum(newid()))*(10)+20,2)
Here the random number will come in between 20 and 30.
round will give two decimal place maximum.
If you want negative numbers you can do it with
select round(rand(checksum(newid()))*(10)-60,2)
Then the min value will be -60 and max will be -50.
try using a seed value in the RAND(seedInt). RAND() will only execute once per statement that is why you see the same number each time.
If you don't need it to be an integer, but any random unique identifier, you can use newid()
SELECT table_name, newid() magic_number
FROM information_schema.tables
You would need to call RAND() for each row. Here is a good example
https://web.archive.org/web/20090216200320/http://dotnet.org.za/calmyourself/archive/2007/04/13/sql-rand-trap-same-value-per-row.aspx
The problem I sometimes have with the selected "Answer" is that the distribution isn't always even. If you need a very even distribution of random 1 - 14 among lots of rows, you can do something like this (my database has 511 tables, so this works. If you have less rows than you do random number span, this does not work well):
SELECT table_name, ntile(14) over(order by newId()) randomNumber
FROM information_schema.tables
This kind of does the opposite of normal random solutions in the sense that it keeps the numbers sequenced and randomizes the other column.
Remember, I have 511 tables in my database (which is pertinent only b/c we're selecting from the information_schema). If I take the previous query and put it into a temp table #X, and then run this query on the resulting data:
select randomNumber, count(*) ct from #X
group by randomNumber
I get this result, showing me that my random number is VERY evenly distributed among the many rows:
It's as easy as:
DECLARE #rv FLOAT;
SELECT #rv = rand();
And this will put a random number between 0-99 into a table:
CREATE TABLE R
(
Number int
)
DECLARE #rv FLOAT;
SELECT #rv = rand();
INSERT INTO dbo.R
(Number)
values((#rv * 100));
SELECT * FROM R
select ABS(CAST(CAST(NEWID() AS VARBINARY) AS INT)) as [Randomizer]
has always worked for me
Use newid()
select newid()
or possibly this
select binary_checksum(newid())
If you want to generate a random number between 1 and 14 inclusive.
SELECT CONVERT(int, RAND() * (14 - 1) + 1)
OR
SELECT ABS(CHECKSUM(NewId())) % (14 -1) + 1
DROP VIEW IF EXISTS vwGetNewNumber;
GO
Create View vwGetNewNumber
as
Select CAST(RAND(CHECKSUM(NEWID())) * 62 as INT) + 1 as NextID,
'abcdefghijklmnopqrstuvwxyz0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ'as alpha_num;
---------------CTDE_GENERATE_PUBLIC_KEY -----------------
DROP FUNCTION IF EXISTS CTDE_GENERATE_PUBLIC_KEY;
GO
create function CTDE_GENERATE_PUBLIC_KEY()
RETURNS NVARCHAR(32)
AS
BEGIN
DECLARE #private_key NVARCHAR(32);
set #private_key = dbo.CTDE_GENERATE_32_BIT_KEY();
return #private_key;
END;
go
---------------CTDE_GENERATE_32_BIT_KEY -----------------
DROP FUNCTION IF EXISTS CTDE_GENERATE_32_BIT_KEY;
GO
CREATE function CTDE_GENERATE_32_BIT_KEY()
RETURNS NVARCHAR(32)
AS
BEGIN
DECLARE #public_key NVARCHAR(32);
DECLARE #alpha_num NVARCHAR(62);
DECLARE #start_index INT = 0;
DECLARE #i INT = 0;
select top 1 #alpha_num = alpha_num from vwGetNewNumber;
WHILE #i < 32
BEGIN
select top 1 #start_index = NextID from vwGetNewNumber;
set #public_key = concat (substring(#alpha_num,#start_index,1),#public_key);
set #i = #i + 1;
END;
return #public_key;
END;
select dbo.CTDE_GENERATE_PUBLIC_KEY() public_key;
Update my_table set my_field = CEILING((RAND(CAST(NEWID() AS varbinary)) * 10))
Number between 1 and 10.
Try this:
SELECT RAND(convert(varbinary, newid()))*(b-a)+a magic_number
Where a is the lower number and b is the upper number
If you need a specific number of random number you can use recursive CTE:
;WITH A AS (
SELECT 1 X, RAND() R
UNION ALL
SELECT X + 1, RAND(R*100000) --Change the seed
FROM A
WHERE X < 1000 --How many random numbers you need
)
SELECT
X
, RAND_BETWEEN_1_AND_14 = FLOOR(R * 14 + 1)
FROM A
OPTION (MAXRECURSION 0) --If you need more than 100 numbers