I'm trying to link two tables, one has an 'EntityRef' that's made of four alpha characters and a sequential number...
EntityRef
=========
SWIT1
LIVE32
KIRB48
MEHM38
BRAD192
The table that I'm trying to link to stores the reference in a 15 character field where the 4 alphas are at the start and the numbers are at the end but with zeros in between to make up the 15 characters...
EntityRef
=========
SWIT00000000001
LIVE00000000032
So, to get theses to link, my options are to either remove the zeros on one field or add the zeros on the other.
I've gone for the later as it seems to be a simpler approach and eliminates the risk of getting into problems if the numeric element contains a zero.
So, the alpha is always 4 characters at the beginning and the number is the remainder and 15 minus the LEN() of the EntityRef is the number of zeros that I need to insert...
left(entityref,4) as 'Alpha',
right(entityref,len(EntityRef)-4) as 'Numeric',
15-len(EntityRef) as 'No.of Zeros'
Alpha Numeric No.of Zeros
===== ======= ===========
SWIT 1 10
LIVE 32 9
KIRB 48 9
MEHM 38 9
MALL 36 9
So, I need to concatenate the three elements but I don't know how to create the string of zeros to the specified length...how do I do that??
Concat(Alpha, '0'*[No. of Zeros], Numeric)
What is the correct way to repeat a character a specified number of times?
You can use string manipulation. In this case:
LEFT() to get the alpha portion.
REPLICATE() to get the zeros.
STUFF() to get the number.
The query:
select left(val, 4) + replicate('0', 15 - len(val)) + stuff(val, 1, 4, '')
from (values ('SWIT1'), ('ABC12345')) v(val)
You may try left padding with zeroes:
SELECT
LEFT(EntityRef, 4) +
RIGHT('00000000000' + SUBSTRING(ISNULL(EntityRef,''), 5, 30), 11) AS EntityRef
FROM yourTable;
Demo
With casting to integer the numeric part:
select *
from t1 inner join t2
on concat(left(t2.EntityRef, 4), cast(right(t2.EntityRef, 11) as bigint)) = t1.EntityRef
See the demo.
I found the answer as soon as I posted the question (sometimes it helps you think it through!).
(left(entityref,4) + replicate('0',15-len(EntityRef)) +
right(entityref,len(EntityRef)-4)),
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)
ID MOBILE
1 9869600733
2 9869600793
3 9869600799
all id whose mobile number containing 9 three times(using string functions like replace, substr, etc)... ? (without like , % , etc)
You can use LEN and Replace
Where len(MOBILE)-len(replace(MOBILE ,'9',''))>=3
Note : Some DBMS uses LENGTH instead of LEN
Where length(MOBILE)-length(replace(MOBILE ,'9',''))>=3
DEMO
replace(MOBILE ,'9','') will replace all the 9's with empty
string
length(MOBILE) will count the number of characters in Mobile
column
length(replace(MOBILE ,'9','')) will count the number of characters
in Mobile column as replacing 9's with empty string
length(MOBILE)-length(replace(MOBILE ,'9','')) here the
difference will tell the number of missing characters that is our 9, you can use this difference to count the 9
exactly three '9's:
Select * from mytable
Where len(mobile) - len(replace(mobile, '9', '')) = 3
at least three '9's:
Select * from mytable
Where len(mobile) - len(replace(mobile, '9', '')) >= 3
I have a substring which pulls the data I need but I need to add a leading zero to the front of the result. I've searched and found several samples of pulling data with leading zeros but none using a substring. I can add the zero to the end of the result but do not know how to add it to the front. The substring I am using is shown below.
"substring((substring(convert(char(6),Convert(int,a.nor_ppd_hrs_no*100) + 100000),3,4) + space(16)),1,16) as 'ApprovedHrs',"
This produces a result like this 7500 and I need it to look like this 07500.
Thanks
One of most simple solution for adding leading zeros in order to get a number with a desired width (ex. 16 digits) is to concatenate a string of zeros (REPLICATE('0', 15) or '000000000000000') with the source number (ex. 123) converted to VARCHAR ('123'). Then, the result '000000000000000123' is truncated to desired length (ex. 16 digits) using the RIGHT function:
DECLARE #Num INT;
SET #Num = 123;
SELECT RIGHT(REPLICATE('0', 15) + CONVERT(VARCHAR(11), #Num), 16) AS NumWithLeadingZeros1
SELECT RIGHT('000000000000000' + CONVERT(VARCHAR(11), #Num), 16) AS NumWithLeadingZeros2
Output:
NumWithLeadingZeros1
--------------------
0000000000000123
NumWithLeadingZeros2
--------------------
0000000000000123
Assuming you need the resulting string to be 5 characters length,
stuff(substring((substring(convert(char(6),Convert(int,a.nor_ppd_hrs_no*100) + 100000),3,4) + space(16)),1,16), 1, 0, replicate('0', 5 - len(substring((substring(convert(char(6),Convert(int,a.nor_ppd_hrs_no*100) + 100000),3,4) + space(16)),1,16)))) as 'ApprovedHrs
Use the STR() function to convert a number to a fixed-length, right-justified string. By default, it pads with spaces, so replace ' ' with '0' to get a zero-padded output.
SELECT REPLACE(STR(a.nor_ppd_hrs_no*100,5),' ','0') FROM MyTable
I need to update the first digit of numbers.
For example, 3003.
I only want the first '3' to be changed to '2' and don't want the last digit '3' to be changed.
Something like the following faulty query:
update table
set defaulttopicorder = replace(defaulttopicorder, left(defaulttopicorder,1), 2)
where ....
Assuming the defaulttopicorder column is a non-decimal, this will add 1 to the first digit of the number:
SET defaulttopicorder = defaulttopicorder + POWER(10, LEN(STR(defaulttopicorder)))
...so if you want to subtract 1 from the first digit:
SET defaulttopicorder = defaulttopicorder + -1 * POWER(10, LEN(STR(defaulttopicorder)))
Use STR(defaulttopicorder) to make a string out of the number, SUBSTRING around to take the part of it starting at index 2, + to concatenate the leading '2' with that substring.
This sounds like homework, so rather than giving the answer, how about a suggestion.
You can turn the number into a string, then just modify the first character in the varchar, but, when you increase it, remember that it may be incremented if you go from 9 to 10.
Another approach is to someone remove all the other digits, so if you have value:
3976, then have it be 3000, and increment then add back what you had removed.
Here's a quick inelegant approach:
DECLARE #MyInt INT = 5003
PRINT CAST(SUBSTRING(CAST(#MyInt AS VARCHAR(4)), 1, 1) + 1 AS CHAR(1)) + SUBSTRING(CAST(#MyInt AS VARCHAR(4)), 2, LEN(#MyInt) - 1)
Convert to an update as required. Tested on SQL08.
Here's a slightly different approach, which doesn't involve turning it into a string (so it may be more efficient, although you'd have to test it to be certain).
If you have a number N, then log10(N) is the power that 10 would have to be raised to in order to get N.
But if you round it down and raise 10 to that power (i.e. 10floor(log10(N))) you'll get the place value of the leftmost digit.
Now, multiply that place value by the amount you want to add or subtract from the digit (e.g. -1 or 1) and add the result to the original number.
So, given your example:
N = 3003
log10(N) = approx. 3.477
floor(3.477) = 3
103 = 1000
So, if you want to add 1 to the first digit, add 1 * 1000 (=1000) to the number; if you want to subtract 1, add -1 * 1000 (=-1000) to the number.
Of course, you'll want to be careful of rounding errors, which can pop up unexpectedly due to problems with representing decimal values as binary.