This question already has answers here:
Formatting Numbers by padding with leading zeros in SQL Server
(14 answers)
Closed 7 years ago.
I have been working on a query (in sql Server TSQL) which fills left of a number with 0's so output is always 5 digit.
So:
Select MuNumber From Mytable
for data 11,011,2132,1111
Creates output like
00011
02134
01111
I tried Lpad Function but numer of 0's can be different.
if Munumber is 1 we need 0000 and If MyNumber is 34 we need 000
Assuming that MuNumber is VARCHAR simply use RIGHT
SELECT RIGHT('00000' + MuNumber, 5)
FROM Mytable
Otherwise you need to convert it first
SELECT RIGHT('00000' + CONVERT(VARCHAR(5), MuNumber), 5)
FROM Mytable
And in general you can use this pattern:
DECLARE #num INT = 10;
SELECT RIGHT(REPLICATE('0', #num) + CONVERT(VARCHAR(5), MuNumber), #num)
FROM Mytable
Try this
select right('00000'+cast(col as varchar(5)),5) from table
You can use the user defined function udfLeftSQLPadding where you can find the source codes at SQL Pad Leading Zeros
After you create the function on your database, you can use it as follows
select
dbo.udfLeftSQLPadding(MuNumber,5,'0')
from dbo.Mytable
Another option:
declare #n int = 6
select stuff(replicate('0', #n), 6-len(n), len(n), n)
from (values('123'), ('2493'), ('35')) as x(n)
Related
This may sound like a weird request, but I have to make an excel sheet with EXACTLY the same format as their old sheet. The values in the column in question will be a numeric code. I need to display up to 8 digits, and no less than 6 digits. The code will only be 7 or 8 digits if the first number in the sequence is not zero. If there are 6 digits or less, I need to display 6 digits including leading zeroes. Here is an example:
The data comes in like this:
000023547612
000000873901
000031765429
000000000941
000000055701
I need those numbers to display as:
23547612
873901
31765429
000941
055701
Is there a way to achieve this in a SQL statement?
You can check the length of the string and use right to pad it. But GSerg's answer is better.
declare #Test table (Number varchar(12))
insert into #Test (Number)
values ('000023547612'),('000000873901'),('000031765429'),('000000000941'),('000000055701')
select Number
, case when len(convert(varchar(12), convert(int, Number))) <= 6 then right('000000'+convert(varchar(12), convert(int, Number)),6) else convert(varchar(12), convert(int, Number)) end
, format(convert(int, Number), N'##000000') -- GSerg's Answer
from #Test
Returns:
Number Attempt1 Attempt2 (GSerg)
000023547612 23547612 23547612
000000873901 873901 873901
000031765429 31765429 31765429
000000000941 000941 000941
000000055701 055701 055701
PS: In future if you provide test data in this format (table variable or temp table) you will make it much easier for people to answer.
You can use
SELECT Substring('0078956', Patindex('%[^0 ]%', '0078956' + ' '), Len('0078956') ) AS Trimmed_Leading_0;
I have a column with the given values
MRN
1946
456
27
557
The column values length is fixed.
If at all any value is less than 6characters,then it should concate 0's to the left and make it 6characters length.
The desired output is
MRN
001946
000456
000027
000557
This is called left paddings. In SQL Server, this is typically done with more basic string operations:
select right(replicate('0', 6) + mrn, 6)
If mrn is a number, then use the concat() function:
select right(concat(replicate('0', 6), mrn), 6)
You can also use the FORMAT function for this. (Demo)
SELECT FORMAT(MRN ,'D6')
FROM YourTable
Change the number 6 to whatever your total length needs to be:
SELECT REPLICATE('0',6-LEN(EmployeeId)) + EmployeeId
If the column is an INT, you can use RTRIM to implicitly convert it to a VARCHAR
SELECT REPLICATE('0',6-LEN(RTRIM(EmployeeId))) + RTRIM(EmployeeId)
And the code to remove these 0s and get back the 'real' number:
SELECT RIGHT(EmployeeId,(LEN(EmployeeId) - PATINDEX('%[^0]%',EmployeeId)) + 1)
We can achieve this by adding leading zero's
select RIGHT('0000'+CAST(MRN AS VARCHAR(10)),6)
This question already has answers here:
How to count instances of character in SQL Column
(17 answers)
Closed 3 years ago.
Please help me - I have problem with counting how many 'a' (character) there are in a row and column.
This is my query :
declare #zz as varchar(10) = '123a123a12'
select #zz
What function in SQL Server 2008 R2 can I use to count how many 'a' are in there?
How can I combine charindex with len?
Thanks
The old school trick here is to replace the 'a's with empty spaces ('') and compare the resulting string's length to the length of the original:
declare #zz as varchar(10) = '123a123a12'
declare #zz_without_a varchar(10)=replace(#zz,'a','')
declare #a_in_zz int=len(#zz)-len(zz_without_a)
I would use REPLACE and DATALENGTH instead of LEN, because of trailing spaces that could appear after replace.
Example:
DECLARE #zz as varchar(20) = '123a123a12 a'
SELECT DATALENGTH(#zz) - DATALENGTH(REPLACE(#zz, 'a', '')),
LEN(#zz) - LEN(REPLACE(#zz, 'a', ''))
The output is 3 and 9.
If #zz is NVARCHAR, not VARCHAR, you will have to divide by 2.
From MSDN:
DATALENGTH function returns the number of bytes used to represent any
expression
One possibe approach is to use REPLACE() and LEN() functions.:
DECLARE #zz varchar(10) = '123a123a12'
SELECT LEN(#zz) - LEN(REPLACE(#zz, 'a', '')) AS CharCount
Output:
CharCount
2
Another possible approach, if you want to count more than one character, is to use recursion:
DECLARE #zz varchar(10) = 'aa3a123a12'
;WITH cte AS (
SELECT 1 AS N
UNION ALL
SELECT N + 1
FROM cte
WHERE N < LEN(#zz)
)
SELECT COUNT(*) AS CharCount
FROM cte
WHERE SUBSTRING(#zz, N, 1) IN ('a', '1')
So if I have a data (varchar) like say 10.1
I need the value as 0000101000000.
means (000010) whole number and (1000000) decimal value.
Its a 13 character string ,numbers coming before decimal point should be in first 6 characters and numbers coming after decimal point should be in last 7 characters
Maybe..?
DECLARE #d decimal(13,7) = 10.1;
SELECT RIGHT('0000000000000' + CONVERT(varchar(13),CONVERT(bigint,(#d * 10000000))),13);
Using my crystal ball here though.
Edit: As, for some reason, the OP is storing a decimal as a varchar (this is a really bad bad idea on it's own), I have added further logic to attempt to convert the value to a decimal first.
As experience has taught many of us, give a user a non-numeric column to store a numeric value in and they're more than happily store a non-numeric value in it, so i have used TRY_CONVERT and assumed you are using SQL Server 2012+:
DECLARE #d varchar(13) = 10.1;
SELECT RIGHT('0000000000000' + CONVERT(varchar(13),CONVERT(bigint,(TRY_CONVERT(decimal(13,7),#d) * 10000000))),13);
SELECT REPLICATE('0',6-LEN(SUBSTRING(CAST([data] AS VARCHAR), 1,
CHARINDEX('.',CAST([data] AS VARCHAR)) -1)))+SUBSTRING(CAST([data] AS VARCHAR), 1,
CHARINDEX('.',CAST([data] AS VARCHAR)) -1)+
SUBSTRING(CAST([data] AS VARCHAR), CHARINDEX('.',CAST([data] AS VARCHAR)) + 1,
LEN(CAST([data] AS VARCHAR)))+REPLICATE('0',7-LEN(SUBSTRING(CAST([data] AS VARCHAR), CHARINDEX('.',CAST([data] AS VARCHAR)) + 1,
LEN(CAST([data] AS VARCHAR))))) AS Whole
FROM Table1
Output
Whole
0000101000000
Demo
http://sqlfiddle.com/#!18/8649d/16
You can use some math and string operations to do it like below
see live demo
declare #var decimal(10,4)
set #var=10.1
select #var,
right(cast(cast(( floor(#var)+ power(10,7)) as int) as varchar(13)),6)
+
cast(cast(((#var- floor(#var)) * power(10,7)) as int) as varchar(13))
There's a fair amount of string manipulation to be done here. I'll step through what I did.
I used a variable for the base number so I could verify different results:
declare #n decimal(9,3) = 10.1
You need 6 spaces left of the decimal and 7 spaces to the right, so I'm doing all the manipulation on a VARCHAR(13). I didn't create a new variable as a VARCHAR because I'm assuming you want to be able to do this conversion in line on the fly, so I'm using that CAST over and over again.
Start by finding the decimal place.
SELECT CHARINDEX('.',CAST(#n as VARCHAR(13)))
In the sample number, that's a 3, but it could obviously change.
Now, get the portion of the number to the left of the decimal place.
SELECT SUBSTRING(CAST(#n as VARCHAR(13)),1,CHARINDEX('.',CAST(#n as VARCHAR(13)))-1)
Then get the portion to the right of the decimal.
SELECT SUBSTRING(CAST(#n as VARCHAR(13)),CHARINDEX('.',CAST(#n as VARCHAR(13)))+1,LEN(CAST(#n as VARCHAR(13))))
Pad the leading zeroes. Put 6 on, concatenate, and take a RIGHT 6. Accounts for no digits to the left of the decimal.
SELECT RIGHT(REPLICATE(0,6) + SUBSTRING(CAST(#n as VARCHAR(13)),1,CHARINDEX('.',CAST(#n as VARCHAR(13)))-1), 6)
Pad the trailing zeroes. Same idea, but in the other direction.
SELECT LEFT(SUBSTRING(CAST(#n as VARCHAR(13)),CHARINDEX('.',CAST(#n as VARCHAR(13)))+1,LEN(CAST(#n as VARCHAR(13)))) + REPLICATE(0,7),7)
Then put it all together.
SELECT RIGHT(REPLICATE(0,6) + SUBSTRING(CAST(#n as VARCHAR(13)),1,CHARINDEX('.',CAST(#n as VARCHAR(13)))-1), 6)
+
LEFT(SUBSTRING(CAST(#n as VARCHAR(13)),CHARINDEX('.',CAST(#n as VARCHAR(13)))+1,LEN(CAST(#n as VARCHAR(13)))) + REPLICATE(0,7),7)
Results.
0000101000000
declare #var varchar(20) = '10000.112'
SELECT FORMAT (FLOOR(#var), '000000') + left((PARSENAME(#var,1)) + replicate('0',7),7)
I was wondering if anyone could help with a query to select part of a column.
The column 'criteriadata' contains data that would look like this:
CriteriaData
14 27 15 C
14 30 15 DD
14 38 15 Pass
14 33 15 Pass
How can I select just the data that appears after the number 15.
Many thanks.
SELECT RIGHT(CriteriaData,
LEN(CriteriaData) - CHARINDEX('15', CriteriaData, 1) - 2)
FROM TableName
WHERE CriteriaData LIKE '%15%';
SQL Fiddle Demo
declare #T table
(
CriteriaData varchar(20)
)
insert into #T values
('14 27 15 C'),
('14 30 15 DD'),
('14 38 15 Pass'),
('14 33 15 Pass')
select stuff(CriteriaData, 1, 3+charindex(' 15 ', CriteriaData), '')
from #T
Result:
---------
C
DD
Pass
Pass
If CriteriaCData always contains a pattern of 3 numbers of 2 numerics separated by a space then you always want to retrieve from 10th chars:
select SUBSTR(CriteriaCData, 10) from xxx
If you are under oracle min 10.g then use REGEXP_SUBSTR to retrieve the alpha pattern
SELECT upper(REGEXP_SUBSTR(CriteriaCData, '[a-zA-Z]*$')) FROM xxx
Since you seem to want everything from the ninth character onwards, you could use RIGHT and LEN
SELECT right([CriteriaData], len([CriteriaData]) - 9)
However, you'd be better off normalizing your data so it was already in a seperate column.
On oracle use LENGTH instead of LEN
SELECT substr(CriteriaData, 8, LENGTH(CriteriaData) - 9) from table
You should use substring with left functions
Have a look at this: How to extract this specific substring in SQL Server?
And this: http://msdn.microsoft.com/en-us/library/aa259342(v=sql.80).aspx
SELECT substring(criteriadata, 9, LEN(criteriadata)-8) from table
This assumes that the position of 15 is fixed.
Declare #x nvarchar(100) = '14 30 15 DD';
Select substring(#x, (select charindex('15',#x,1) + 2) ,len(#x));
I created a SQL function to split the criteria by the spaces and used the last remaining value after the last space.
create function dbo.getCriteria
(
#criteria varchar(500)
)
returns varchar(500)
begin
declare #space as int
select #space=charindex(' ', data) from mydata
while #space > 0
begin
set #criteria=substring(#criteria, #space + 1, len(#criteria))
select #space=charindex(' ', #criteria)
end
return #criteria
end
select dbo.getCriteria(data) from mydata
SELECT
RIGHT(CriteriaData, LEN(CriteriaData) - (CHARINDEX('15', CriteriaData, 1) - 2))
FROM
MyTable;
As I had trouble making prior answers work, I had to find my own and figure for future reference I'd leave it on Stack Overflow. My field has XML but it's an NVarchar field and should generalise just fine - if you have a clear criteria for left AND right surrounding strings.
It's not a complete match to this question but I hope it helps someone else who has huge strings in their columns and needs to snip out a string that varies in between two others!
WITH r
AS (
SELECT TOP 100 RIGHT(XMLData, LEN(XMLData)-CHARINDEX('<INVOICE_NO>', XMLData)-11) AS xmldata
FROM IncomingPartsInvoiceXML)
SELECT LEFT(xmldata, CHARINDEX('<\/INVOICE_NO>', XMLData)-1)
FROM r;