Select query to remove non-numeric characters - tsql

Question

I've got dirty data in a column with variable alpha length. I just want to strip out anything that is not 0-9.
I do not want to run a function or proc. I have a script that is similar that just grabs the numeric value after text, it looks like this:
Update TableName
set ColumntoUpdate=cast(replace(Columnofdirtydata,'Alpha #','') as int)
where Columnofdirtydata like 'Alpha #%'
And ColumntoUpdate is Null
I thought it would work pretty good until I found that some of the data fields I thought would just be in the format Alpha # 12345789 are not.
Examples of data that needs to be stripped
AB ABCDE # 123
ABCDE# 123
AB: ABC# 123
I just want the 123. It is true that all data fields do have the # prior to the number.
I tried substring and PatIndex, but I'm not quite getting the syntax correct or something. Anyone have any advice on the best way to address this?

Answer 1

See this blog post on extracting numbers from strings in SQL Server. Below is a sample using a string in your example:
DECLARE #textval NVARCHAR(30)
SET #textval = 'AB ABCDE # 123'
SELECT LEFT(SUBSTRING(#textval, PATINDEX('%[0-9.-]%', #textval), 8000),
PATINDEX('%[^0-9.-]%', SUBSTRING(#textval, PATINDEX('%[0-9.-]%', #textval), 8000) + 'X') -1)

Answer 2

Here is an elegant solution if your server supports the TRANSLATE function (on sql server it's available on sql server 2017+ and also sql azure).
First, it replaces any non numeric characters with a # character.
Then, it removes all # characters.
You may need to add additional characters that you know may be present in the second parameter of the TRANSLATE call.
select REPLACE(TRANSLATE([Col], 'abcdefghijklmnopqrstuvwxyz+()- ,#+', '##################################'), '#', '')

Answer 3

You can use stuff and patindex.
stuff(Col, 1, patindex('%[0-9]%', Col)-1, '')
SQL Fiddle

Answer 4

This works well for me:
CREATE FUNCTION [dbo].[StripNonNumerics]
(
#Temp varchar(255)
)
RETURNS varchar(255)
AS
Begin
Declare #KeepValues as varchar(50)
Set #KeepValues = '%[^0-9]%'
While PatIndex(#KeepValues, #Temp) > 0
Set #Temp = Stuff(#Temp, PatIndex(#KeepValues, #Temp), 1, '')
Return #Temp
End
Then call the function like so to see the original something next to the sanitized something:
SELECT Something, dbo.StripNonNumerics(Something) FROM TableA

Answer 5

In case if there are some characters possible between digits (e.g. thousands separators), you may try following:
declare #table table (DirtyCol varchar(100))
insert into #table values
('AB ABCDE # 123')
,('ABCDE# 123')
,('AB: ABC# 123')
,('AB#')
,('AB # 1 000 000')
,('AB # 1`234`567')
,('AB # (9)(876)(543)')
;with tally as (select top (100) N=row_number() over (order by ##spid) from sys.all_columns),
data as (
select DirtyCol, Col
from #table
cross apply (
select (select C + ''
from (select N, substring(DirtyCol, N, 1) C from tally where N<=datalength(DirtyCol)) [1]
where C between '0' and '9'
order by N
for xml path(''))
) p (Col)
where p.Col is not NULL
)
select DirtyCol, cast(Col as int) IntCol
from data
Output is:
DirtyCol IntCol
--------------------- -------
AB ABCDE # 123 123
ABCDE# 123 123
AB: ABC# 123 123
AB # 1 000 000 1000000
AB # 1`234`567 1234567
AB # (9)(876)(543) 9876543
For update, add ColToUpdate to select list of the data cte:
;with num as (...),
data as (
select ColToUpdate, /*DirtyCol, */Col
from ...
)
update data
set ColToUpdate = cast(Col as int)

Answer 6

CREATE FUNCTION FN_RemoveNonNumeric (#Input NVARCHAR(512))
RETURNS NVARCHAR(512)
AS
BEGIN
DECLARE #Trimmed NVARCHAR(512)
SELECT #Trimmed = #Input
WHILE PATINDEX('%[^0-9]%', #Trimmed) > 0
SELECT #Trimmed = REPLACE(#Trimmed, SUBSTRING(#Trimmed, PATINDEX('%[^0-9]%', #Trimmed), 1), '')
RETURN #Trimmed
END
GO
SELECT dbo.FN_RemoveNonNumeric('ABCDE# 123')

Answer 7

Pretty late to the party, I found the following which I though worked brilliantialy.. if anyone is still looking
SELECT
(SELECT CAST(CAST((
SELECT SUBSTRING(FieldToStrip, Number, 1)
FROM master..spt_values
WHERE Type='p' AND Number <= LEN(FieldToStrip) AND
SUBSTRING(FieldToStrip, Number, 1) LIKE '[0-9]' FOR XML Path(''))
AS xml) AS varchar(MAX)))
FROM
SourceTable

Answer 8

Here's a version which pulls all digits from a string; i.e. given I'm 35 years old; I was born in 1982. The average family has 2.4 children. this would return 35198224. i.e. it's good where you've got numeric data which may have been formatted as a code (e.g. #123,456,789 / 123-00005), but isn't appropriate if you're looking to pull out specific numbers (i.e. as opposed to digits / just the numeric characters) from the text. Also it only handles digits; so won't return negative signs (-) or periods .).
declare #table table (id bigint not null identity (1,1), data nvarchar(max))
insert #table (data)
values ('hello 123 its 45613 then') --outputs: 12345613
,('1 some other string 98 example 4') --outputs: 1984
,('AB ABCDE # 123') --outputs: 123
,('ABCDE# 123') --outputs: 123
,('AB: ABC# 123') --outputs: 123
; with NonNumerics as (
select id
, data original
--the below line replaces all digits with blanks
, replace(replace(replace(replace(replace(replace(replace(replace(replace(replace(data,'0',''),'1',''),'2',''),'3',''),'4',''),'5',''),'6',''),'7',''),'8',''),'9','') nonNumeric
from #table
)
--each iteration of the below CTE removes another non-numeric character from the original string, putting the result into the numerics column
, Numerics as (
select id
, replace(original, substring(nonNumeric,1,1), '') numerics
, replace(nonNumeric, substring(nonNumeric,1,1), '') charsToreplace
, len(replace(nonNumeric, substring(nonNumeric,1,1), '')) charsRemaining
from NonNumerics
union all
select id
, replace(numerics, substring(charsToreplace,1,1), '') numerics
, replace(charsToreplace, substring(charsToreplace,1,1), '') charsToreplace
, len(replace(charsToreplace, substring(charsToreplace,1,1), '')) charsRemaining
from Numerics
where charsRemaining > 0
)
--we select only those strings with `charsRemaining=0`; i.e. the rows for which all non-numeric characters have been removed; there should be 1 row returned for every 1 row in the original data set.
select * from Numerics where charsRemaining = 0
This code works by removing all the digits (i.e. the characters we want) from a the given strings by replacing them with blanks. Then it goes through the original string (which includes the digits) removing all of the characters that were left (i.e. the non-numeric characters), thus leaving only the digits.
The reason we do this in 2 steps, rather than just removing all non-numeric characters in the first place is there are only 10 digits, whilst there are a huge number of possible characters; so replacing that small list is relatively fast; then gives us a list of those non-numeric characters which actually exist in the string, so we can then replace that small set.
The method makes use of recursive SQL, using common table expressions (CTEs).

Answer 9

To add on to Ken's answer, this handles commas and spaces and parentheses
--Handles parentheses, commas, spaces, hyphens..
declare #table table (c varchar(256))
insert into #table
values
('This is a test 111-222-3344'),
('Some Sample Text (111)-222-3344'),
('Hello there 111222 3344 / How are you?'),
('Hello there 111 222 3344 ? How are you?'),
('Hello there 111 222 3344. How are you?')
select
replace(LEFT(SUBSTRING(replace(replace(replace(replace(replace(c,'(',''),')',''),'-',''),' ',''),',',''), PATINDEX('%[0-9.-]%', replace(replace(replace(replace(replace(c,'(',''),')',''),'-',''),' ',''),',','')), 8000),
PATINDEX('%[^0-9.-]%', SUBSTRING(replace(replace(replace(replace(replace(c,'(',''),')',''),'-',''),' ',''),',',''), PATINDEX('%[0-9.-]%', replace(replace(replace(replace(replace(c,'(',''),')',''),'-',''),' ',''),',','')), 8000) + 'X') -1),'.','')
from #table

Answer 10

Create function fn_GetNumbersOnly(#pn varchar(100))
Returns varchar(max)
AS
BEGIN
Declare #r varchar(max) ='', #len int ,#c char(1), #x int = 0
Select #len = len(#pn)
while #x <= #len
begin
Select #c = SUBSTRING(#pn,#x,1)
if ISNUMERIC(#c) = 1 and #c <> '-'
Select #r = #r + #c
Select #x = #x +1
end
return #r
End

Answer 11

In your case It seems like the # will always be after teh # symbol so using CHARINDEX() with LTRIM() and RTRIM() would probably perform the best. But here is an interesting method of getting rid of ANY non digit. It utilizes a tally table and table of digits to limit which characters are accepted then XML technique to concatenate back to a single string without the non-numeric characters. The neat thing about this technique is it could be expanded to included ANY Allowed characters and strip out anything that is not allowed.
DECLARE #ExampleData AS TABLE (Col VARCHAR(100))
INSERT INTO #ExampleData (Col) VALUES ('AB ABCDE # 123'),('ABCDE# 123'),('AB: ABC# 123')
DECLARE #Digits AS TABLE (D CHAR(1))
INSERT INTO #Digits (D) VALUES ('0'),('1'),('2'),('3'),('4'),('5'),('6'),('7'),('8'),('9')
;WITH cteTally AS (
SELECT
I = ROW_NUMBER() OVER (ORDER BY (SELECT NULL))
FROM
#Digits d10
CROSS APPLY #Digits d100
--add more cross applies to cover longer fields this handles 100
)
SELECT *
FROM
#ExampleData e
OUTER APPLY (
SELECT CleansedPhone = CAST((
SELECT TOP 100
SUBSTRING(e.Col,t.I,1)
FROM
cteTally t
INNER JOIN #Digits d
ON SUBSTRING(e.Col,t.I,1) = d.D
WHERE
I <= LEN(e.Col)
ORDER BY
t.I
FOR XML PATH('')) AS VARCHAR(100))) o

Answer 12

Declare #MainTable table(id int identity(1,1),TextField varchar(100))
INSERT INTO #MainTable (TextField)
VALUES
('6B32E')
declare #i int=1
Declare #originalWord varchar(100)=''
WHile #i<=(Select count(*) from #MainTable)
BEGIN
Select #originalWord=TextField from #MainTable where id=#i
Declare #r varchar(max) ='', #len int ,#c char(1), #x int = 0
Select #len = len(#originalWord)
declare #pn varchar(100)=#originalWord
while #x <= #len
begin
Select #c = SUBSTRING(#pn,#x,1)
if(#c!='')
BEGIN
if ISNUMERIC(#c) = 0 and #c <> '-'
BEGIN
Select #r = cast(#r as varchar) + cast(replace((SELECT ASCII(#c)-64),'-','') as varchar)
end
ELSE
BEGIN
Select #r = #r + #c
END
END
Select #x = #x +1
END
Select #r
Set #i=#i+1
END

Answer 13

I have created a function for this
Create FUNCTION RemoveCharacters (#text varchar(30))
RETURNS VARCHAR(30)
AS
BEGIN
declare #index as int
declare #newtexval as varchar(30)
set #index = (select PATINDEX('%[A-Z.-/?]%', #text))
if (#index =0)
begin
return #text
end
else
begin
set #newtexval = (select STUFF ( #text , #index , 1 , '' ))
return dbo.RemoveCharacters(#newtexval)
end
return 0
END
GO

Answer 14

Here is the answer:
DECLARE #t TABLE (tVal VARCHAR(100))
INSERT INTO #t VALUES('123')
INSERT INTO #t VALUES('123S')
INSERT INTO #t VALUES('A123,123')
INSERT INTO #t VALUES('a123..A123')
;WITH cte (original, tVal, n)
AS
(
SELECT t.tVal AS original,
LOWER(t.tVal) AS tVal,
65 AS n
FROM #t AS t
UNION ALL
SELECT tVal AS original,
CAST(REPLACE(LOWER(tVal), LOWER(CHAR(n)), '') AS VARCHAR(100)),
n + 1
FROM cte
WHERE n <= 90
)
SELECT t1.tVal AS OldVal,
t.tval AS NewVal
FROM (
SELECT original,
tVal,
ROW_NUMBER() OVER(PARTITION BY tVal + original ORDER BY original) AS Sl
FROM cte
WHERE PATINDEX('%[a-z]%', tVal) = 0
) t
INNER JOIN #t t1
ON t.original = t1.tVal
WHERE t.sl = 1

Answer 15

You can create SQL CLR scalar function in order to be able to use regular expressions like replace patterns.
Here you can find example of how to create such function.
Having such function will solve the issue with just the following lines:
SELECT [dbo].[fn_Utils_RegexReplace] ('AB ABCDE # 123', '[^0-9]', '');
SELECT [dbo].[fn_Utils_RegexReplace] ('ABCDE# 123', '[^0-9]', '');
SELECT [dbo].[fn_Utils_RegexReplace] ('AB: ABC# 123', '[^0-9]', '');
More important, you will be able to solve more complex issues as the regular expressions will bring a whole new world of options directly in your T-SQL statements.

Answer 16

Use this:
REPLACE(TRANSLATE(SomeString, REPLACE(TRANSLATE(SomeString, '0123456789', '##########'), '#', ''), REPLICATE('#', LEN(REPLACE(TRANSLATE(SomeString, '0123456789', '##########'), '#', '') + 'x') - 1)), '#', '')
Demo:
DROP TABLE IF EXISTS #MyTempTable;
CREATE TABLE #MyTempTable (SomeString VARCHAR(255));
INSERT INTO #MyTempTable
VALUES ('ssss123ssg99d362sdg')
, ('hey 62q&*^(n43')
, (NULL)
, ('')
, ('hi')
, ('123');
SELECT SomeString
, REPLACE(TRANSLATE(SomeString, REPLACE(TRANSLATE(SomeString, '0123456789', '##########'), '#', ''), REPLICATE('#', LEN(REPLACE(TRANSLATE(SomeString, '0123456789', '##########'), '#', '') + 'x') - 1)), '#', '')
FROM #MyTempTable;
DROP TABLE IF EXISTS #MyTempTable;
Results:
SomeString
(No column name)
ssss123ssg99d362sdg
12399362
hey62q&*^(n43
6243
NULL
NULL
hi
123
123

Answer 17

While the OP wanted to "strip out anything that is not 0-9", the post is also tagged with "substring" and "patindex", and the OP mentioned the concern "not quite getting the syntax correct or something". To address that the requirements note that "all data fields do have the # prior to the number" and to provide an answer that addresses the challenges with substring/patindex, consider the following:
/* A sample select */
;WITH SampleValues AS
( SELECT 'AB ABCDE # 123' [Columnofdirtydata]
UNION ALL SELECT 'AB2: ABC# 123')
SELECT
s.Columnofdirtydata,
f1.pos1,
'['+ f2.substr +']' [InspectOutput]
FROM
SampleValues s
CROSS APPLY (SELECT PATINDEX('%# %',s.Columnofdirtydata) [pos1]) f1
CROSS APPLY (SELECT SUBSTRING(s.Columnofdirtydata, f1.pos1 + LEN('#-'),LEN(s.Columnofdirtydata)) [substr]) f2
/* Using update scenario from OP */
UPDATE t1
SET t1.Columntoupdate = CAST(f2.substr AS INT)
FROM
TableName t1
CROSS APPLY (SELECT PATINDEX('%# %',t1.Columnofdirtydata) [pos1]) f1
CROSS APPLY (SELECT SUBSTRING(t1.Columnofdirtydata, f1.pos1 + LEN('#-'),LEN(t1.Columnofdirtydata)) [substr]) f2
Note that my syntax advice for patindex/substring, is to:
consider using APPLY as a way to temporarily alias results from one function for use as parameters in the next. It's not uncommon to (in ETL, for example) need to parse out parameter/position-based substrings in an updatable column of a staging table. If you need to "debug" and potentially fix some parsing logic, this style will help.
consider using LEN('PatternSample') in your substring logic, to account for reusing this pattern or adjusting it when your source data changes (instead of "+ 1"
SUBSTRING() requires a length parameter, but it can be greater than the length of the string. Therefore, if you are getting "the rest of the string" after the pattern, you can just use "The source length"

Answer 18

DECLARE #STR VARCHAR(400)
DECLARE #specialchars VARCHAR(50) = '%[~,#,#,$,%,&,*,(,),!^?:]%'
SET #STR = '1, 45 4,3 68.00-'
WHILE PATINDEX( #specialchars, #STR ) > 0
---Remove special characters using Replace function
SET #STR = Replace(Replace(REPLACE( #STR, SUBSTRING( #STR, PATINDEX( #specialchars, #STR ), 1 ),''),'-',''), ' ','')
SELECT #STR

Answer 19

SELECT REGEXP_REPLACE( col, '[^[:digit:]]', '' ) AS new_col FROM my_table