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
Related
I am trying to filter out some query results to where it only shows items with 6 decimal places. I don't need it to round up or add 0's to the answer, just filter out anything that is 5 decimal places or below. My current query looks like this: (ex. if item is 199.54215 i dont want to see it but if it is 145.253146 i need it returned)
select
TRA_CODPLANTA,
TRA_WO,
TRA_IMASTER,
tra_codtipotransaccion,
tra_Correlativo,
TRA_INGRESOFECHA,
abs(tra_cantidadparcial) as QTY
from mw_tra_transaccion
where FLOOR (Tra_cantidadparcial*100000) !=tra_cantidadparcial*100000
and substring(tra_imaster,1,2) not in ('CP','SG','PI','MR')
and TRA_CODPLANTA not in ('4Q' , '5C' , '5V' , '8H' , '7W' , 'BD', 'DP')
AND tra_INGRESOFECHA > #from_date
and abs(tra_cantidadparcial) > 0.00000
Any assistance would be greatly appreciated!
Here is an example with ROUND, which seems to be the ideal function to use, since it remains in the realms of numbers. If you have at most 5 decimal places, then rounding to 5 decimal places will leave the value unchanged.
create table #test (Tra_cantidadparcial decimal(20,10));
INSERT #test (Tra_cantidadparcial) VALUES (1),(99999.999999), (1.000001), (45.000001), (45.00001);
SELECT * FROM #test WHERE ROUND(Tra_cantidadparcial,5) != Tra_cantidadparcial;
drop table #test
If your database values are VARCHAR and exist in the DB like so:
100.123456
100.1
100.100
You can achieve this using a wildcard LIKE statement example
WHERE YOUR_COLUMN_NAME LIKE '%.[0-9][0-9][0-9][0-9][0-9][0-9]%'
This will being anything containing a decimal place followed by AT LEAST 6 numeric values
Here is an example using a conversion to varchar and using the LEN - the CHARINDEX of the decimal point, I'm not saying this is the best way, but you did ask for an example in syntax, so here you go:
--Temp Decimal value holding up to 10 decimal places and 10 whole number places
DECLARE #temp DECIMAL(20, 10) = 123.4565432135
--LEN returns an integer number of characters in the converted varchar
--CHARINDEX returns the integer location of the decimal where it is found in the varchar
--IF the number of characters left after subtracting the index of the decimal from the length of the varchar is greater than 5,
--you have more than 6 decimal places
IF LEN(CAST(#temp AS varchar(20))) - CHARINDEX('.', CAST(#temp AS varchar(20)), 0) > 5
SELECT 1
ELSE
SELECT 0
Here is a shorthand way.
WHERE (LEN(CONVERT(DOUBLE PRECISION, FieldName % 1)) - 2) >=5
One way would be to convert / cast that column to a lower precision. Doing this would cause automatic rounding, but that would show you if it is 6 decimals or not based on the last digit. If the last digit of the converted value is 0, then it's false, otherwise it's true.
declare #table table (v decimal(11,10))
insert into #table
values
(1.123456789),
(1.123456),
(1.123),
(1.123405678)
select
v
,cast(v as decimal(11,5)) --here, we are changing the value to have a precision of 5. Notice the rounding.
,right(cast(v as decimal(11,5)),1) --this is taking the last digit. If it's 0, we don't want it
from #table
Thus, your where clause would simply be.
where right(cast(tra_cantidadparcial as decimal(11,5)),1) > 0
I am trying to get last numeric part in the given string.
For Example, below are the given strings and the result should be last numeric part only
SB124197 --> 124197
287276ACBX92 --> 92
R009321743-16 --> 16
How to achieve this functionality. Please help.
Try this:
select right(#str, patindex('%[^0-9]%',reverse(#str)) - 1)
Explanation:
Using PATINDEX with '%[^0-9]%' as a search pattern you get the starting position of the first occurrence of a character that is not a number.
Using REVERSE you get the position of the first non numeric character starting from the back of the string.
Edit:
To handle the case of strings not containing non numeric characters you can use:
select case
when patindex(#str, '%[^0-9]%') = 0 then #str
else right(#str, patindex('%[^0-9]%',reverse(#str)) - 1)
end
If your data always contains at least one non-numeric character then you can use the first query, otherwise use the second one.
Actual query:
So, if your table is something like this:
mycol
--------------
SB124197
287276ACBX92
R009321743-16
123456
then you can use the following query (works in SQL Server 2012+):
select iif(x.i = 0, mycol, right(mycol, x.i - 1))
from mytable
cross apply (select patindex('%[^0-9]%', reverse(mycol) )) as x(i)
Output:
mynum
------
124197
92
16
123456
Demo here
Here is one way using Patindex
SELECT RIGHT(strg, COALESCE(NULLIF(Patindex('%[^0-9]%', Reverse(strg)), 0) - 1, Len(strg)))
FROM (VALUES ('SB124197'),
('287276ACBX92'),
('R009321743-16')) tc (strg)
After reversing the string, we are finding the position of first non numeric character and extracting the data from that position till the end..
Result :
-----
124197
92
16
Have decimals stored as varchar.
I have a column with value 0.0375000. I need to convert this into 0.0375.
When I did
convert(decimal(8, 7), substring(column, 0, 1) + '.' + substring(column, 2, 8)))
I got the result as 0.0375000.
I want to remove all the trailing zeros and the result I want is 0.0375
How can I do this?
If 2012+ The #'s indicate an optional display
Select format(0.0375000,'0.######') Returns 0.0375
Select format(0.037502,'0.######') Returns 0.037502
Sorry didn't see stored as varchar()
Select format(cast(somecolumn as decimal(18,8)),'0.######')
if you only need 4 decimal places, you want decimal 5,4 (assuming your number to the left of the decimal point fits into 1 digit , if you need 2 digits, choose decimal(6,4) for example )
select convert(decimal(5,4), substring(column,0,1)+'.' +substring(column,2,8) )
decimal data type https://msdn.microsoft.com/en-gb/library/ms187746.aspx
--SQL Code for easy removing trailing zeros.
select CONVERT(DOUBLE PRECISION,'2.256000')
--Result will be 2.256
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 in a SQL Server table that has the following rows:
MyColumn : C1_xxx1,C2_xxx1,C3_xxx1,C1_xxx2,C1_xxx3,C3_xxx2 etc
It is a text column that contains strings that have the following format: CY_mystring where Y is a number from 1 to 5, followed by the '_' character then mystring that can have any value.
Is there a way to make a select return this column ordered as following:
C1_xxx1
C1_xxx2
C1_xxx3
......
C1_xxxn
C2_xxx1
......
C2_xxxn
C3_xxx1
.......
C3_xxxn
etc
Ordered by the CY_ substring.
thank you
This should do it .. (order first by the first two chars, and then by the last char (assuming that the final n is always one digit long))
SELECT
Column1
FROM
TABLENAME
ORDER BY
LEFT(Column1,2) ASC,
RIGHT(Column1,1) ASC
You say that Y is a number from 1 to 5 it's always one character long. Assuming the format is xY_xxxZ, you can order on Y then Z like:
order by
substring(MyColumn,2,1) -- Second digit
, right(MyColumn,1) -- Last digit
If Z can be longer than one character (i.e. 10 or higher) you can use pathindex to determine the number of digits at the end:
order by
substring(MyColumn,2,1) -- Second digit
, right(MyColumn, patindex('%[^0-9]%', reverse(MyColumn))-1) -- Digits at end