split alpha and numeric using sql - sql-server

Question

I have a table and it has a 3 columns. The first column is the data that contains value(numeric) and unit(percentage and etc..), the second column is numeric column, the third is Unit column. What I want to do is split the numeric and the unit from the first column then put those split-ted data to its designated column.
Here is my table:
I tried this function:SO link here..., it really does splitting alpha and numeric but then I'm new in using SQL Function, my problem there is the parameter must be in string STRING, so what I did is change it to Sub Query but it gives me error.
Sample COde:
SQL FUNCTION:
create function [dbo].[GetNumbersFromText](#String varchar(2000))
returns table as return
(
with C as
(
select cast(substring(S.Value, S1.Pos, S2.L) as int) as Number,
stuff(s.Value, 1, S1.Pos + S2.L, '') as Value
from (select #String+' ') as S(Value)
cross apply (select patindex('%[0-9]%', S.Value)) as S1(Pos)
cross apply (select patindex('%[^0-9]%', stuff(S.Value, 1, S1.Pos, ''))) as S2(L)
union all
select cast(substring(S.Value, S1.Pos, S2.L) as int),
stuff(S.Value, 1, S1.Pos + S2.L, '')
from C as S
cross apply (select patindex('%[0-9]%', S.Value)) as S1(Pos)
cross apply (select patindex('%[^0-9]%', stuff(S.Value, 1, S1.Pos, ''))) as S2(L)
where patindex('%[0-9]%', S.Value) > 0
)
select Number
from C
)
SELECT STATEMENT with SUB Query:
declare #S varchar(max)
select number from GetNumbersFromText(Select SomeColm From Table_Name) option (maxrecursion 0)
BTW, im using sql server 2005.
Thanks!

Answer 1

If the numeric part is always at the beginning, then you can use this:
PATINDEX('%[0-9][^0-9]%', ConcUnit)
to get the index of the last digit.
Thus, this:
DECLARE #str VARCHAR(MAX) = '4000 ug/ML'
SELECT LEFT(#str, PATINDEX('%[0-9][^0-9]%', #str )) AS Number,
LTRIM(RIGHT(#str, LEN(#str) - PATINDEX('%[0-9][^0-9]%', #str ))) As Unit
gives you:
Number Unit
-------------
4000 ug/ML
EDIT:
If numeric data include double values as well, then you can use this:
SELECT LEN(#str) - PATINDEX ('%[^0-9][0-9]%', REVERSE(#str))
to get the index of the last digit.
Thus, this:
SELECT LEFT(#str, LEN(#str) - PATINDEX ('%[^0-9][0-9]%', REVERSE(#str)))
gives you the numeric part.
And this:
SELECT LEFT(#str, LEN(#str) - PATINDEX ('%[^0-9][0-9]%', REVERSE(#str))) AS Numeric,
CASE
WHEN CHARINDEX ('%', #str) <> 0 THEN LTRIM(RIGHT(#str, LEN(#str) - CHARINDEX ('%', #str)))
ELSE LTRIM(RIGHT(#str, PATINDEX ('%[^0-9][0-9]%', REVERSE(#str))))
END AS Unit
gives you both numberic and unit part.
Here are some tests that I made with the data you have posted:
Input:
DECLARE #str VARCHAR(MAX) = '50 000ug/ML'
Output:
Numeric Unit
------------
50 000 ug/ML
Input:
DECLARE #str VARCHAR(MAX) = '99.5%'
Output:
Numeric Unit
------------
99.5
Input:
DECLARE #str VARCHAR(MAX) = '4000 . 35 % ug/ML'
Output:
Numeric Unit
------------------
4000 . 35 ug/ML

Answer 2

Here is my answer. Check output in SQLFiddle for the same.
create TABLE temp
(
string NVARCHAR(50)
)
INSERT INTO temp (string)
VALUES
('4000 ug\ml'),
('2000 ug\ml'),
('%'),
('ug\ml')
SELECT subsrtunit,LEFT(subsrtnumeric, PATINDEX('%[^0-9]%', subsrtnumeric+'t') - 1)
FROM (
SELECT subsrtunit = SUBSTRING(string, posofchar, LEN(string)),
subsrtnumeric = SUBSTRING(string, posofnumber, LEN(string))
FROM (
SELECT string, posofchar = PATINDEX('%[^0-9]%', string),
posofnumber = PATINDEX('%[0-9]%', string)
FROM temp
) d
) t
Updated Version to handle 99.5 ug\ml
create TABLE temp
(
string NVARCHAR(50)
)
INSERT INTO temp (string)
VALUES
('4000 ug\ml'),
('2000 ug\ml'),
('%'),
('ug\ml'),
('99.5 ug\ml')
SELECT subsrtunit,LEFT(subsrtnumeric, PATINDEX('%[^0-9.]%', subsrtnumeric+'t') - 1)
FROM (
SELECT subsrtunit = SUBSTRING(string, posofchar, LEN(string)),
subsrtnumeric = SUBSTRING(string, posofnumber, LEN(string))
FROM (
SELECT string, posofchar = PATINDEX('%[^0-9.]%', string),
posofnumber = PATINDEX('%[0-9.]%', string)
FROM temp
) d
) t
Updated Version: To handle 1 000 ug\ml,20 000ug\ml
create TABLE temp
(
string NVARCHAR(50)
)
INSERT INTO temp (string)
VALUES
('4000 ug\ml'),
('2000 ug\ml'),
('%'),
('ug\ml'),
('99.5 ug\ml'),
('1 000 ug\ml'),
('20 000ug\ml')
SELECT substring(replace(subsrtunit,' ',''),PATINDEX('%[0-9.]%', replace(subsrtunit,' ',''))+1,len(subsrtunit)),
LEFT(replace(subsrtnumeric,' ',''), PATINDEX('%[^0-9.]%', replace(subsrtnumeric,' ','')+'t') - 1)
FROM (
SELECT subsrtunit = SUBSTRING(string, posofchar, LEN(string)),
subsrtnumeric = SUBSTRING(string, posofnumber, LEN(string))
FROM (
SELECT string, posofchar = PATINDEX('%[^0-9.]%', replace(string,' ','')),
posofnumber = PATINDEX('%[0-9.]%', replace(string,' ',''))
FROM temp
) d
) t
Check out SQLFiddle for the same.

Answer 3

Would something like this work? Based on the shown data it looks like it would.
Apply it to your data set as a select and if you like the results then you can make an update from it.
WITH cte as (SELECT 'ug/mL' ConcUnit, 500 as [Numeric], '' as Unit
UNION ALL SELECT '2000 ug/mL', NULL, '')
SELECT
[ConcUnit] as [ConcUnit],
[Numeric] as [Original Numeric],
[Unit] as [Original Unit],
CASE WHEN ConcUnit LIKE '% %' THEN
SUBSTRING(ConcUnit, 1, CHARINDEX(' ', ConcUnit) - 1)
ELSE [Numeric] END as [New Numeric],
CASE WHEN ConcUnit LIKE '% %'
THEN SUBSTRING(ConcUnit, CHARINDEX(' ', ConcUnit) + 1, LEN(ConcUnit))
ELSE ConcUnit END as [New Unit]
FROM cte

Answer 4

change #concunit & #unitx Respectively
DECLARE #concunit varchar(10)='45.5%'
DECLARE #unitx varchar(10)='%'
BEGIN
SELECT RTRIM(SUBSTRING( #concunit , 1 , CHARINDEX( #unitx , #concunit
) - 1
)) AS Number,
RTRIM(SUBSTRING( #concunit , CHARINDEX( #unitx , #concunit
) , LEN( #concunit
) - (CHARINDEX( #unitx , #concunit
) - 1)
)) AS Unit
end

Answer 5

I had the same dilemma, but in my case the alpha's were in front of the numerics.
So using the logic that #Giorgos Betsos added to his answer, I just reversed it.
I.e., when your input is :
abc123
You can split it like this:
declare #input varchar(30) = 'abc123'
select
replace(#input,reverse(LEFT(reverse(#input), PATINDEX('%[0-9][^0-9]%', reverse(#input) ))),'') Alpha
, reverse(LEFT(reverse(#input), PATINDEX('%[0-9][^0-9]%', reverse(#input) ))) Numeric
Results :