I need to return two fields from a database concatenated as 'field1-field2'. The second field is an int, but needs to be returned as a fixed length of 5 with leading 0's. The method i'm using is:
SELECT Field1 + '-' + RIGHT('0000' + CAST(Field2 AS varchar),5) FROM ...
Is there a more efficient way to do this?
That is pretty much the way: Adding Leading Zeros To Integer Values
So, to save following the link, the query looks like this, where #Numbers is the table and Num is the column:
SELECT RIGHT('000000000' + CONVERT(VARCHAR(8),Num), 8) FROM #Numbers
for negative or positive values
declare #v varchar(6)
select #v = -5
SELECT case when #v < 0
then '-' else '' end + RIGHT('00000' + replace(#v,'-',''), 5)
Another way (without CAST or CONVERT):
SELECT RIGHT(REPLACE(STR(#NUM),' ','0'),5)
If you can afford/want to have a function in your database you could use something like:
CREATE FUNCTION LEFTPAD
(#SourceString VARCHAR(MAX),
#FinalLength INT,
#PadChar CHAR(1))
RETURNS VARCHAR(MAX)
AS
BEGIN
RETURN
(SELECT Replicate(#PadChar, #FinalLength - Len(#SourceString)) + #SourceString)
END
I would do it like this.
SELECT RIGHT(REPLICATE('0', 5) + CAST(Field2 AS VARCHAR(5),5)
Not necessarily all that "Easier", or more efficient, but better to read. Could be optimized to remove the need for "RIGHT"
If you want to get a consistent number of total strings in the final result by adding different number of zeros, here is a little bit modification (for vsql)
SELECT
CONCAT(
REPEAT('0', 9-length(TO_CHAR(var1))),
CAST(var1 AS VARCHAR(9))
) as var1
You can replace 9 by any number for your need!
BRD
Related
How can I remove value before '_' and show date and time in one row in TSQL Function?
Below is sample:
Declare #inputstring as varchar(50) = 'Studio9_20230126_203052' ;
select value from STRING_SPLIT( #inputstring ,'_')
Output Required: 2023-01-26 20:30:52.000
If we can safely assume that the value is always in the format {Some String}_{yyyyMMdd}_{hhmmss} then you can use STUFF a few times, firstly to remove the leading string up to the first underscore (_) character (using CHARINDEX to find that character), and then to inject 2 colon (:) characters. Finally you can REPLACE the remaining underscore with a space ( ), and then use TRY_CONVERT to attempt to convert the value to a datetime2(0).
DECLARE #inputstring varchar(50) = 'Studio9_20230126_203052';
SELECT TRY_CONVERT(datetime2(0),REPLACE(STUFF(STUFF(STUFF(#inputstring,1,CHARINDEX('_',#inputstring),''),14,0,':'),12,0,':'),'_',' '));
Note that this doesn't give the value you state you want in your question (2023-01-26 20:05:52.000) , but I assume this is a typographical error, and that the 05 for minutes should be 30.
Creating function
CREATE FUNCTION [dbo].[convert_to_date] (#inputstring NVARCHAR(MAX))
RETURNS DATETIME AS
BEGIN
DECLARE #finalString varchar(50), #out varchar(100)
SET #finalString = REPLACE ( (SUBSTRING (#inputstring, CHARINDEX('_', #inputstring)+1 , LEN(#inputstring))), '_', ' ')
--SELECT #finalString
SET #out = LEFT (#finalString, 4) + '-'
+ SUBSTRING(#finalString, 5, 2) + '-'
+ SUBSTRING(#finalString, 7, 2) + ' '
+ SUBSTRING(#finalString, 10, 2) + ':'
+ SUBSTRING(#finalString, 12, 2) + ':'
+ SUBSTRING(#finalString, 14, 2) + '.000'
RETURN #out
END
Select Query
SELECT dbo.[convert_to_date] ('Studio54541659_20230126_203052')
Output
2023-01-26 20:30:52.000
This will tolerate "somestring" in the format of "somestring_YYYYMMDD_HHMISS" being variable in length.
Declare #inputstring as varchar(50) = 'Studio9_20230126_203052' ;
SELECT DateAndTime = CONVERT(DATETIME,STUFF(STUFF(STUFF(v2.DT,14,0,':'),12,0,':'),9,1,' '))
,Identifier = LEFT(#inputstring,v1.Pos1-1) --Included this because I know how people are :D --Comment out if not wanted.
,Original = #inputstring --Original string just for checking. Comment out when happy.
FROM (VALUES(CHARINDEX('_',#inputstring)))v1(Pos1) --Position of first Underscore
CROSS APPLY (VALUES(SUBSTRING(#inputstring,v1.Pos1+1,50)))v2(DT) --String after first Underscore
;
Output looks like this and you end up with a DATETIME datatype. Comment out what you don't want for columns in the return.
I'll let you have some of the fun by converting it into an iTVF (inline Table Valued Function). Remember that any function that contains a "BEGIN" is ultimately going to be a part of a performance issue so make sure it's an iTVF :D
EDIT: Crud... I've gotta remember to scroll down. #Lamu already posted the same thing but it's probably better and fast if you just want the time and not the identifier I included.
I have the following code in a stored procedure and am trying to conditionally format a calculated number based on its length (if the number is less than 4 digits, pad with leading zeros). However, my case statement is not working. The "formattedNumber2" result is the one I'm looking for.
I'm assuming the case statement treats the variable strangely, but I also don't know of a way around this.
DECLARE #Number int = 5
SELECT
CASE
WHEN (LEN(CONVERT(VARCHAR, #Number)) > 4)
THEN #Number
ELSE RIGHT('0000' + CAST(#Number AS VARCHAR(4)), 4)
END AS formattedNumber,
LEN(CONVERT(VARCHAR, #Number)) AS numberLength,
RIGHT('0000' + CAST(#Number AS VARCHAR(4)), 4) AS formattedNumber2
I get the following results when I run the query:
formattedNumber numberLength formattedNumber2
-------------------------------------------------
5 1 0005
SQL DEMO
The problem is you are using different data type on your case , integer and string. So the CASE stay with the first type he find and convert the rest.
CASE WHEN (LEN(convert(VARCHAR, #Number)) > 4) THEN convert(VARCHAR, #Number)
This can be done a lot easier with format() since version 2012.
format(n,
'0000')
And that would also handle negative values, which your current approach apparently doesn't.
Prior 2012 it can be handled with basically replicate() and + (string concatenation).
isnull(replicate('-',
-sign(n)), '')
+
isnull(replicate('0',
4
-
len(cast(abs(n) AS varchar(10)))
),
'')
+
cast(abs(n) AS varchar(10))
(It targets integer values, choose a larger length for the varchar casts for bigint.)
db<>fiddle
I have a numeric field like 1,3065 and I need that to became like this: 000000000000000130.
I mean 16 integers and 2 decimals, without the comma, adding 0es to the left and, if needed, to the right.
Is there any way to do that with a query?
I think this will work in Sybase:
select right(replicate('0', 16) + cast(cast(field*100 as int) as varchar(255)), 16)
Try this way:
select REPLICATE('0',16-len(cast(cast((1.3065*100) as int) as varchar(16))))+cast(cast((1.3065*100) as int) as varchar(16))
Result of above select:
0000000000000130
If you want approximation you should use FLOOR or CEILING function as below
declare #multiplier int
select #multiplier = 1000
select REPLICATE('0',16-len(cast((floor(1.3065*#multiplier)) as varchar(16))))+cast(floor(1.3065*#multiplier) as varchar(16))
Result of above select:
0000000000001306
You can change an approximation by changing #multiplier. For example if you want result from 1st query you should change #multiplier to 100.
Is there a way in T-SQL to convert a TINYINT to VARCHAR with custom number formatting?
For instance, my TINYINT has a value of 3 and I want to convert it to a VARCH of 03, so that it always shows a 2 digit number.
I don't see this ability in the CONVERT function.
RIGHT('00' + CONVERT(VARCHAR, MyNumber), 2)
Be warned that this will cripple numbers > 99. You might want to factor in that possibility.
Use the RIGHT function...
e.g.
DECLARE #testnum TINYINT
SET #testnum = 3
PRINT RIGHT('00' + CONVERT(VARCHAR(2), #testnum), 2)
You can try this
DECLARE #Table TABLE(
Val INT
)
INSERT INTO #Table SELECT 3
INSERT INTO #Table SELECT 30
DECLARE #NumberPrefix INT
SET #NumberPrefix = 2
SELECT REPLICATE('0', #NumberPrefix - LEN(Val)) + CAST(Val AS VARCHAR(10))
FROM #Table
What is the value range? Is it 0 through 10? If so, then try:
SELECT REPLICATE('0',2-LEN(#t)) + CAST(#t AS VARCHAR)
That handles 0 through 9 as well as 10 through 99.
Now, tinyint can go up to the value of 255. If you want to handle > 99 through 255, then try this solution:
declare #t TINYINT
set #t =233
SELECT ISNULL(REPLICATE('0',2-LEN(#t)),'') + CAST(#t AS VARCHAR)
To understand the solution, the expression to the left of the + calculates the number of zeros to prefix to the string.
In case of the value 3, the length is 1. 2 - 1 is 1. REPLICATE Adds one zero.
In case of the value 10, the length is 2. 2 - 2 is 0. REPLICATE Adds nothing.
In the case of the value 100, the length is -1 which produces a NULL. However, the null value is handled and set to an empty string.
Now if you decide that because tinyint can contain up to 255 and you want your formatting as three characters, just change the 2-LEN to 3-LEN in the left expression and you're set.
declare #t tinyint
set #t =3
select right(replicate('0', 2) + cast(#t as varchar),2)
Ditto: on the cripping effect for numbers > 99
If you want to cater for 1-255 then you could use
select right(replicate('0', 2) + cast(#t as varchar),3)
But this would give you 001, 010, 100 etc
Here's an alternative following the last answer
declare #t tinyint,#v tinyint
set #t=23
set #v=232
Select replace(str(#t,4),' ','0'),replace(str(#t,5),' ','0')
This will work on any number and by varying the length of the str() function you can stipulate how many leading zeros you require. Provided of course that your string length is always >= maximum number of digits your number type can hold.
Correción: 3-LEN
declare #t TINYINT
set #t =233
SELECT ISNULL(REPLICATE('0',3-LEN(#t)),'') + CAST(#t AS VARCHAR)
Had the same problem with a zipcode field. Some folks sent me an excel file with zips, but they were formatted as #'s. Had to convert them to strings as well as prepend leading 0's to them if they were < 5 len ...
declare #int tinyint
set #int = 25
declare #len tinyint
set #len = 3
select right(replicate('0', #len) + cast(#int as varchar(255)), #len)
You just alter the #len to get what you want. As formatted, you'll get...
001
002
...
010
011
...
255
Ideally you'd "varchar(#len)", too, but that blows up the SQL compile. Have to toss an actual # into it instead of a var.
I am working on a SQL query that reads from a SQLServer database to produce an extract file. One of the requirements to remove the leading zeroes from a particular field, which is a simple VARCHAR(10) field. So, for example, if the field contains '00001A', the SELECT statement needs to return the data as '1A'.
Is there a way in SQL to easily remove the leading zeroes in this way? I know there is an RTRIM function, but this seems only to remove spaces.
select substring(ColumnName, patindex('%[^0]%',ColumnName), 10)
select replace(ltrim(replace(ColumnName,'0',' ')),' ','0')
You can use this:
SELECT REPLACE(LTRIM(REPLACE('000010A', '0', ' ')),' ', '0')
I had the same need and used this:
select
case
when left(column,1) = '0'
then right(column, (len(column)-1))
else column
end
select substring(substring('B10000N0Z', patindex('%[0]%','B10000N0Z'), 20),
patindex('%[^0]%',substring('B10000N0Z', patindex('%[0]%','B10000N0Z'),
20)), 20)
returns N0Z, that is, will get rid of leading zeroes and anything that comes before them.
If you want the query to return a 0 instead of a string of zeroes or any other value for that matter you can turn this into a case statement like this:
select CASE
WHEN ColumnName = substring(ColumnName, patindex('%[^0]%',ColumnName), 10)
THEN '0'
ELSE substring(ColumnName, patindex('%[^0]%',ColumnName), 10)
END
In case you want to remove the leading zeros from a string with a unknown size.
You may consider using the STUFF command.
Here is an example of how it would work.
SELECT ISNULL(STUFF(ColumnName
,1
,patindex('%[^0]%',ColumnName)-1
,'')
,REPLACE(ColumnName,'0','')
)
See in fiddler various scenarios it will cover
https://dbfiddle.uk/?rdbms=sqlserver_2012&fiddle=14c2dca84aa28f2a7a1fac59c9412d48
You can try this - it takes special care to only remove leading zeroes if needed:
DECLARE #LeadingZeros VARCHAR(10) ='-000987000'
SET #LeadingZeros =
CASE WHEN PATINDEX('%-0', #LeadingZeros) = 1 THEN
#LeadingZeros
ELSE
CAST(CAST(#LeadingZeros AS INT) AS VARCHAR(10))
END
SELECT #LeadingZeros
Or you can simply call
CAST(CAST(#LeadingZeros AS INT) AS VARCHAR(10))
Here is the SQL scalar value function that removes leading zeros from string:
SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO
-- =============================================
-- Author: Vikas Patel
-- Create date: 01/31/2019
-- Description: Remove leading zeros from string
-- =============================================
CREATE FUNCTION dbo.funRemoveLeadingZeros
(
-- Add the parameters for the function here
#Input varchar(max)
)
RETURNS varchar(max)
AS
BEGIN
-- Declare the return variable here
DECLARE #Result varchar(max)
-- Add the T-SQL statements to compute the return value here
SET #Result = #Input
WHILE LEFT(#Result, 1) = '0'
BEGIN
SET #Result = SUBSTRING(#Result, 2, LEN(#Result) - 1)
END
-- Return the result of the function
RETURN #Result
END
GO
To remove the leading 0 from month following statement will definitely work.
SELECT replace(left(Convert(nvarchar,GETDATE(),101),2),'0','')+RIGHT(Convert(nvarchar,GETDATE(),101),8)
Just Replace GETDATE() with the date field of your Table.
To remove leading 0, You can multiply number column with 1
Eg: Select (ColumnName * 1)
select CASE
WHEN TRY_CONVERT(bigint,Mtrl_Nbr) = 0
THEN ''
ELSE substring(Mtrl_Nbr, patindex('%[^0]%',Mtrl_Nbr), 18)
END
you can try this
SELECT REPLACE(columnname,'0','') FROM table
I borrowed from ideas above. This is neither fast nor elegant. but it is accurate.
CASE
WHEN left(column, 3) = '000' THEN right(column, (len(column)-3))
WHEN left(column, 2) = '00' THEN right(a.column, (len(column)-2))
WHEN left(column, 1) = '0' THEN right(a.column, (len(column)-1))
ELSE
END
select ltrim('000045', '0') from dual;
LTRIM
-----
45
This should do.