I have a SQL entry that is of type hex (varbinary) and I want to do a SELECT COUNT for all the entries that have this hex value ending in 1.
I was thinking about using CONVERT to make my hex into a char and then use WHERE my_string LIKE "%1". The thing is that varchar is capped at 8000 chars, and my hex is longer than that.
What options do I have?
Varbinary actually works with some string manipulation functions, most notably substring. So you can use eg.:
select substring(yourBinary, 1, 1);
To get the first byte of your binary column. To get the last bit then, you can use this:
select substring(yourBinary, len(yourBinary), 1) & 1;
This will give you zero if the bit is off, or one if it is on.
However, if you really only have to check at most the last 4-8 bytes, you can easily use the bitwise operators on the column directly:
select yourBinary & 1;
As a final note, this is going to be rather slow. So if you plan on doing this often, on large amounts of data, it might be better to simply create another bit column just for that, which you can index. If you're talking about at most a thousand rows or so, or if you don't care about speed, fire away :)
Check last four bits = 0001
SELECT SUM(CASE WHEN MyColumn % 16 IN (-15,1) THEN 1 END) FROM MyTable
Check last bit = 1
SELECT SUM(CASE WHEN MyColumn % 2 IN (-1,1) THEN 1 END) FROM MyTable
If you are wondering why you have to check for negative moduli, try SELECT 0x80000001 % 16
Try using this where
WHERE LEFT(my_string,1) = 1
It it's text values ending in 1 then you want the Right as opposed to the Left
WHERE RIGHT(my_string,1) = 1
Related
I have a column named Number
Suppose a value is 12345678997
and I want to output as *******8997, that means all leading digits will be masked by * except the last 4 digits
how can I achieve this using SQL Server Select ?
Try this:
declare #i bigint = '12345678997'
select stuff(#i,1,len(#i)-4,'*******')
Output:
*******8997
Using REPLICATE you can generate a string with a given number of the same character.
Then just add the last four digits to that.
select
Number,
concat(replicate('*',len(Number)-4), right(Number,4)) as MaskedNumber
from YourTable
I need to get numbers with a variable length out of a string and sum them.
The strings got the following format:
EH:NUMBER=SomeOtherStuff->Code
I'm extracting the code via RIGHT() and join with another table to get the group right, at the moment I'm using sum to get it together via date:
SUM(CASE WHEN (MONTH(data.DATE1) = 5 AND YEAR(data.DATE1) = YEAR(GETDATE())) THEN 1 ELSE 0 END) N'Mai',
I then need to sum the numbers from the string and not the number of rows.
Some Examples:
Month1 EH:1=24->ZTM
Month1 EH:4=13-21->LKm
Month2 EH:3=34,33,43->LKm
Month2 EH:7=12,92-29,29->LKm
Month2 EH:5=24-26,11,21,22->ZOL
What i need:
Material - Month1 - Month2
ZTM - 1 - 0
LKM - 4 - 10
ZOL - 0 - 5
Could you help me please?
Greetings
Short version:
What you are looking for is SUBSTRING.
Longer version:
To get the the sum of the numerical value of NUMBER you need think about how break it down.
I'd recommend following these steps:
Extract the NUMBER part from the string. This should be done with SUBSTRING (much like you extract Code with RIGHT). To get the start and and length och your substring use charindex ( or patindex if you like).
Convert the NUMBER part to a numerical value with cast (or convert or what you are familiar with)
Now you can do your aggregation.
So SUM(CAST(SUBSTRING(*this part you will have to figure out by yourself)) as correct numerical data type)).
I'll let you figure out the values to insert by yourself and would recommend to first find the positions of the delimiting characters, then extract the NUMBER part, then get the numerical value .... you get it .
This to gain a better understanding of what you are actually doing.
Cheers, and good luck with your assignment
Martin
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 have a column that is typically only numbers (sometimes it's letters, but that's not important).
How can I make it natural sort?
Currently sorts like this: {1,10,11,12,2,3,4,5,6,7,8,9}
I want it to sort like this: {1,2,3,4,5,6,7,8,9,10,11,12}
IsNumeric is "broken", ISNUMERIC(CHAR(13)) returns 1 and CAST will fail.
Use ISNUMERIC(textval + 'e0'). Final code:
ORDER BY
PropertyName,
CASE ISNUMERIC(MixedField + 'e0') WHEN 1 THEN 0 ELSE 1 END, -- letters after numbers
CASE ISNUMERIC(MixedField + 'e0') WHEN 1 THEN CAST(MixedField AS INT) ELSE 0 END,
MixedField
You can mix order parameters...
Cast it. Also, don't forget to use IsNumeric to make sure you only get the numbers back (if they include letters it IS important ;).
SELECT textval FROM tablename
WHERE IsNumeric(textval) = 1
ORDER BY CAST(textval as int)
Also, cast to the datatype that will hold the largest value.
If you need the non-numbers in the result set too then just append a UNION query where IsNumeric = 0 (order by whatever you want) either before or after.
Have you tied using:
'OrderBy ColumnName Asc'
at the end of your query.
I am working with a table that comes from an external source, and cannot be "cleaned". There is a column which an nvarchar(20) and contains an integer about 95% of the time, but occasionally contains an alpha. I want to use something like
select * from sch.tbl order by cast(shouldBeANumber as integer)
but this throws an error on the odd "3A" or "D" or "SUPERCEDED" value.
Is there a way to say "sort it like a number if you can, otherwise just sort by string"? I know there is some sloppiness in that statement, but that is basically what I want.
Lets say for example the values were
7,1,5A,SUPERCEDED,2,5,SECTION
I would be happy if these were sorted in any of the following ways (because I really only need to work with the numeric ones)
1,2,5,7,5A,SECTION,SUPERCEDED
1,2,5,5A,7,SECTION,SUPERCEDED
SECTION,SUPERCEDED,1,2,5,5A,7
5A,SECTION,SUPERCEDED,1,2,5,7
I really only need to work with the
numeric ones
this will give you only the numeric ones, sorted properly:
SELECT
*
FROM YourTable
WHERE ISNUMERIC(YourColumn)=1
ORDER BY YourColumn
select
*
from
sch.tbl
order by
case isnumeric(shouldBeANumber)
when 1 then cast(shouldBeANumber as integer)
else 0
end
Provided that your numbers are not more than 100 characters long:
WITH chars AS
(
SELECT 1 AS c
UNION ALL
SELECT c + 1
FROM chars
WHERE c <= 99
),
rows AS
(
SELECT '1,2,5,7,5A,SECTION,SUPERCEDED' AS mynum
UNION ALL
SELECT '1,2,5,5A,7,SECTION,SUPERCEDED'
UNION ALL
SELECT 'SECTION,SUPERCEDED,1,2,5,5A,7'
UNION ALL
SELECT '5A,SECTION,SUPERCEDED,1,2,5,7'
)
SELECT rows.*
FROM rows
ORDER BY
(
SELECT SUBSTRING(mynum, c, 1) AS [text()]
FROM chars
WHERE SUBSTRING(mynum, c, 1) BETWEEN '0' AND '9'
FOR XML PATH('')
) DESC
SELECT
(CASE ISNUMERIC(shouldBeANumber)
WHEN 1 THEN
RIGHT(CONCAT('00000000',shouldBeANumber), 8)
ELSE
shouoldBeANumber) AS stringSortSafeAlpha
ORDEER BY
stringSortSafeAlpha
This will add leading zeros to all shouldBeANumber values that truly are numbers and leave all remaining values alone. This way, when you sort, you can use an alpha sort but still get the correct values (with an alpha sort, "100" would be less than "50", but if you change "50" to "050", it works fine). Note, for this example, I added 8 leading zeros, but you only need enough leading zeros to cover the largest possible integer in your column.