'Query to get only numbers from a string
I have data like this:
string 1: 003Preliminary Examination Plan
string 2: Coordination005
string 3: Balance1000sheet
The output I expect is
string 1: 003
string 2: 005
string 3: 1000
And I want to implement it in SQL.
Solution 1:[1]
First create this UDF
CREATE FUNCTION dbo.udf_GetNumeric
(
@strAlphaNumeric VARCHAR(256)
)
RETURNS VARCHAR(256)
AS
BEGIN
DECLARE @intAlpha INT
SET @intAlpha = PATINDEX('%[^0-9]%', @strAlphaNumeric)
BEGIN
WHILE @intAlpha > 0
BEGIN
SET @strAlphaNumeric = STUFF(@strAlphaNumeric, @intAlpha, 1, '' )
SET @intAlpha = PATINDEX('%[^0-9]%', @strAlphaNumeric )
END
END
RETURN ISNULL(@strAlphaNumeric,0)
END
GO
Now use the function as
SELECT dbo.udf_GetNumeric(column_name)
from table_name
I hope this solved your problem.
Solution 2:[2]
Try this one -
Query:
DECLARE @temp TABLE
(
string NVARCHAR(50)
)
INSERT INTO @temp (string)
VALUES
('003Preliminary Examination Plan'),
('Coordination005'),
('Balance1000sheet')
SELECT LEFT(subsrt, PATINDEX('%[^0-9]%', subsrt + 't') - 1)
FROM (
SELECT subsrt = SUBSTRING(string, pos, LEN(string))
FROM (
SELECT string, pos = PATINDEX('%[0-9]%', string)
FROM @temp
) d
) t
Output:
----------
003
005
1000
Solution 3:[3]
Query:
DECLARE @temp TABLE
(
string NVARCHAR(50)
)
INSERT INTO @temp (string)
VALUES
('003Preliminary Examination Plan'),
('Coordination005'),
('Balance1000sheet')
SELECT SUBSTRING(string, PATINDEX('%[0-9]%', string), PATINDEX('%[0-9][^0-9]%', string + 't') - PATINDEX('%[0-9]%',
string) + 1) AS Number
FROM @temp
Solution 4:[4]
Please try:
declare @var nvarchar(max)='Balance1000sheet'
SELECT LEFT(Val,PATINDEX('%[^0-9]%', Val+'a')-1) from(
SELECT SUBSTRING(@var, PATINDEX('%[0-9]%', @var), LEN(@var)) Val
)x
Solution 5:[5]
Getting only numbers from a string can be done in a one-liner. Try this :
SUBSTRING('your-string-here', PATINDEX('%[0-9]%', 'your-string-here'), LEN('your-string-here'))
NB: Only works for the first int in the string, ex: abc123vfg34 returns 123.
Solution 6:[6]
With the previous queries I get these results:
'AAAA1234BBBB3333' >>>> Output: 1234
'-çã+0!\aº1234' >>>> Output: 0
The code below returns All numeric chars:
1st output: 12343333
2nd output: 01234
declare @StringAlphaNum varchar(255)
declare @Character varchar
declare @SizeStringAlfaNumerica int
declare @CountCharacter int
set @StringAlphaNum = 'AAAA1234BBBB3333'
set @SizeStringAlfaNumerica = len(@StringAlphaNum)
set @CountCharacter = 1
while isnumeric(@StringAlphaNum) = 0
begin
while @CountCharacter < @SizeStringAlfaNumerica
begin
if substring(@StringAlphaNum,@CountCharacter,1) not like '[0-9]%'
begin
set @Character = substring(@StringAlphaNum,@CountCharacter,1)
set @StringAlphaNum = replace(@StringAlphaNum, @Character, '')
end
set @CountCharacter = @CountCharacter + 1
end
set @CountCharacter = 0
end
select @StringAlphaNum
Solution 7:[7]
declare @puvodni nvarchar(20)
set @puvodni = N'abc1d8e8ttr987avc'
WHILE PATINDEX('%[^0-9]%', @puvodni) > 0 SET @puvodni = REPLACE(@puvodni, SUBSTRING(@puvodni, PATINDEX('%[^0-9]%', @puvodni), 1), '' )
SELECT @puvodni
Solution 8:[8]
I did not have rights to create functions but had text like
["blahblah012345679"]
And needed to extract the numbers out of the middle
Note this assumes the numbers are grouped together and not at the start and end of the string.
select substring(column_name,patindex('%[0-9]%', column_name),patindex('%[0-9][^0-9]%', column_name)-patindex('%[0-9]%', column_name)+1)
from table name
Solution 9:[9]
This UDF will work for all types of strings:
CREATE FUNCTION udf_getNumbersFromString (@string varchar(max))
RETURNS varchar(max)
AS
BEGIN
WHILE @String like '%[^0-9]%'
SET @String = REPLACE(@String, SUBSTRING(@String, PATINDEX('%[^0-9]%', @String), 1), '')
RETURN @String
END
Solution 10:[10]
I found this approach works about 3x faster than the top voted answer. Create the following function, dbo.GetNumbers:
CREATE FUNCTION dbo.GetNumbers(@String VARCHAR(8000))
RETURNS VARCHAR(8000)
AS
BEGIN;
WITH
Numbers
AS (
--Step 1.
--Get a column of numbers to represent
--every character position in the @String.
SELECT 1 AS Number
UNION ALL
SELECT Number + 1
FROM Numbers
WHERE Number < LEN(@String)
)
,Characters
AS (
SELECT Character
FROM Numbers
CROSS APPLY (
--Step 2.
--Use the column of numbers generated above
--to tell substring which character to extract.
SELECT SUBSTRING(@String, Number, 1) AS Character
) AS c
)
--Step 3.
--Pattern match to return only numbers from the CTE
--and use STRING_AGG to rebuild it into a single string.
SELECT @String = STRING_AGG(Character,'')
FROM Characters
WHERE Character LIKE '[0-9]'
--allows going past the default maximum of 100 loops in the CTE
OPTION (MAXRECURSION 8000)
RETURN @String
END
GO
Testing
Testing for purpose:
SELECT dbo.GetNumbers(InputString) AS Numbers
FROM ( VALUES
('003Preliminary Examination Plan') --output: 003
,('Coordination005') --output: 005
,('Balance1000sheet') --output: 1000
,('(111) 222-3333') --output: 1112223333
,('[email protected]#\-6') --output: 1380046
) testData(InputString)
Testing for performance: Start off setting up the test data...
--Add table to hold test data
CREATE TABLE dbo.NumTest (String VARCHAR(8000))
--Make an 8000 character string with mix of numbers and letters
DECLARE @Num VARCHAR(8000) = REPLICATE('12tf56se',800)
--Add this to the test table 500 times
DECLARE @n INT = 0
WHILE @n < 500
BEGIN
INSERT INTO dbo.NumTest VALUES (@Num)
SET @n = @n +1
END
Now testing the dbo.GetNumbers function:
SELECT dbo.GetNumbers(NumTest.String) AS Numbers
FROM dbo.NumTest -- Time to complete: 1 min 7s
Then testing the UDF from the top voted answer on the same data.
SELECT dbo.udf_GetNumeric(NumTest.String)
FROM dbo.NumTest -- Time to complete: 3 mins 12s
Inspiration for dbo.GetNumbers
Decimals
If you need it to handle decimals, you can use either of the following approaches, I found no noticeable performance differences between them.
- change
'[0-9]'to'[0-9.]' - change
Character LIKE '[0-9]'toISNUMERIC(Character) = 1(SQL treats a single decimal as numeric)
Bonus
You can easily adapt this to differing requirements by swapping out WHERE Character LIKE '[0-9]' with the following options:
WHERE Letter LIKE '[a-zA-Z]' --Get only lettersWHERE Letter LIKE '[0-9a-zA-Z]' --Remove non-alphanumericWHERE Letter LIKE '[^0-9a-zA-Z]' --Get only non-alphanumeric
Solution 11:[11]
Just a little modification to @Epsicron 's answer
SELECT SUBSTRING(string, PATINDEX('%[0-9]%', string), PATINDEX('%[0-9][^0-9]%', string + 't') - PATINDEX('%[0-9]%',
string) + 1) AS Number
FROM (values ('003Preliminary Examination Plan'),
('Coordination005'),
('Balance1000sheet')) as a(string)
no need for a temporary variable
Solution 12:[12]
Although this is an old thread its the first in google search, I came up with a different answer than what came before. This will allow you to pass your criteria for what to keep within a string, whatever that criteria might be. You can put it in a function to call over and over again if you want.
declare @String VARCHAR(MAX) = '-123. a 456-78(90)'
declare @MatchExpression VARCHAR(255) = '%[0-9]%'
declare @return varchar(max)
WHILE PatIndex(@MatchExpression, @String) > 0
begin
set @return = CONCAT(@return, SUBSTRING(@string,patindex(@matchexpression, @string),1))
SET @String = Stuff(@String, PatIndex(@MatchExpression, @String), 1, '')
end
select (@return)
Solution 13:[13]
T-SQL function to read all the integers from text and return the one at the indicated index, starting from left or right, also using a starting search term (optional):
create or alter function dbo.udf_number_from_text(
@text nvarchar(max),
@search_term nvarchar(1000) = N'',
@number_position tinyint = 1,
@rtl bit = 0
) returns int
as
begin
declare @result int = 0;
declare @search_term_index int = 0;
if @text is null or len(@text) = 0 goto exit_label;
set @text = trim(@text);
if len(@text) = len(@search_term) goto exit_label;
if len(@search_term) > 0
begin
set @search_term_index = charindex(@search_term, @text);
if @search_term_index = 0 goto exit_label;
end;
if @search_term_index > 0
if @rtl = 0
set @text = trim(right(@text, len(@text) - @search_term_index - len(@search_term) + 1));
else
set @text = trim(left(@text, @search_term_index - 1));
if len(@text) = 0 goto exit_label;
declare @patt_number nvarchar(10) = '%[0-9]%';
declare @patt_not_number nvarchar(10) = '%[^0-9]%';
declare @number_start int = 1;
declare @number_end int;
declare @found_numbers table (id int identity(1,1), val int);
while @number_start > 0
begin
set @number_start = patindex(@patt_number, @text);
if @number_start > 0
begin
if @number_start = len(@text)
begin
insert into @found_numbers(val)
select cast(substring(@text, @number_start, 1) as int);
break;
end;
else
begin
set @text = right(@text, len(@text) - @number_start + 1);
set @number_end = patindex(@patt_not_number, @text);
if @number_end = 0
begin
insert into @found_numbers(val)
select cast(@text as int);
break;
end;
else
begin
insert into @found_numbers(val)
select cast(left(@text, @number_end - 1) as int);
if @number_end = len(@text)
break;
else
begin
set @text = trim(right(@text, len(@text) - @number_end));
if len(@text) = 0 break;
end;
end;
end;
end;
end;
if @rtl = 0
select @result = coalesce(a.val, 0)
from (select row_number() over (order by m.id asc) as c_row, m.val
from @found_numbers as m) as a
where a.c_row = @number_position;
else
select @result = coalesce(a.val, 0)
from (select row_number() over (order by m.id desc) as c_row, m.val
from @found_numbers as m) as a
where a.c_row = @number_position;
exit_label:
return @result;
end;
Example:
select dbo.udf_number_from text(N'Text text 10 text, 25 term', N'term',2,1);
returns 10;
Solution 14:[14]
Firstly find out the number's starting length then reverse the string to find out the first position again(which will give you end position of number from the end). Now if you deduct 1 from both number and deduct it from string whole length you'll get only number length. Now get the number using SUBSTRING
declare @fieldName nvarchar(100)='AAAA1221.121BBBB'
declare @lenSt int=(select PATINDEX('%[0-9]%', @fieldName)-1)
declare @lenEnd int=(select PATINDEX('%[0-9]%', REVERSE(@fieldName))-1)
select SUBSTRING(@fieldName, PATINDEX('%[0-9]%', @fieldName), (LEN(@fieldName) - @lenSt -@lenEnd))
Solution 15:[15]
This is one of the simplest and easiest one. This will work on the entire String for multiple occurences as well.
CREATE FUNCTION dbo.fn_GetNumbers(@strInput NVARCHAR(500))
RETURNS NVARCHAR(500)
AS
BEGIN
DECLARE @strOut NVARCHAR(500) = '', @intCounter INT = 1
WHILE @intCounter <= LEN(@strInput)
BEGIN
SELECT @strOut = @strOut + CASE WHEN SUBSTRING(@strInput, @intCounter, 1) LIKE '[0-9]' THEN SUBSTRING(@strInput, @intCounter, 1) ELSE '' END
SET @intCounter = @intCounter + 1
END
RETURN @strOut
END Solution 16:[16]
Following a solution using a single common table expression (CTE).
DECLARE @s AS TABLE (id int PRIMARY KEY, value nvarchar(max));
INSERT INTO @s
VALUES
(1, N'003Preliminary Examination Plan'),
(2, N'Coordination005'),
(3, N'Balance1000sheet');
SELECT * FROM @s ORDER BY id;
WITH t AS (
SELECT
id,
1 AS i,
SUBSTRING(value, 1, 1) AS c
FROM
@s
WHERE
LEN(value) > 0
UNION ALL
SELECT
t.id,
t.i + 1 AS i,
SUBSTRING(s.value, t.i + 1, 1) AS c
FROM
t
JOIN @s AS s ON t.id = s.id
WHERE
t.i < LEN(s.value)
)
SELECT
id,
STRING_AGG(c, N'') WITHIN GROUP (ORDER BY i ASC) AS value
FROM
t
WHERE
c LIKE '[0-9]'
GROUP BY
id
ORDER BY
id;
Solution 17:[17]
In Oracle
You can get what you want using this:
SUBSTR('ABCD1234EFGH',REGEXP_INSTR ('ABCD1234EFGH', '[[:digit:]]'),REGEXP_COUNT ('ABCD1234EFGH', '[[:digit:]]'))
Sample Query:
SELECT SUBSTR('003Preliminary Examination Plan ',REGEXP_INSTR ('003Preliminary Examination Plan ', '[[:digit:]]'),REGEXP_COUNT ('003Preliminary Examination Plan ', '[[:digit:]]')) SAMPLE1,
SUBSTR('Coordination005',REGEXP_INSTR ('Coordination005', '[[:digit:]]'),REGEXP_COUNT ('Coordination005', '[[:digit:]]')) SAMPLE2,
SUBSTR('Balance1000sheet',REGEXP_INSTR ('Balance1000sheet', '[[:digit:]]'),REGEXP_COUNT ('Balance1000sheet', '[[:digit:]]')) SAMPLE3 FROM DUAL
Solution 18:[18]
For the hell of it...
This solution is different to all earlier solutions, viz:
- There is no need to create a function
- There is no need to use pattern matching
- There is no need for a temporary table
- This solution uses a recursive common table expression (CTE)
But first - note the question does not specify where such strings are stored. In my solution below, I create a CTE as a quick and dirty way to put these strings into some kind of "source table".
Note also - this solution uses a recursive common table expression (CTE) - so don't get confused by the usage of two CTEs here. The first is simply to make the data avaliable to the solution - but it is only the second CTE that is required in order to solve this problem. You can adapt the code to make this second CTE query your existing table, view, etc.
Lastly - my coding is verbose, trying to use column and CTE names that explain what is going on and you might be able to simplify this solution a little. I've added in a few pseudo phone numbers with some (expected and atypical, as the case may be) formatting for the fun of it.
with SOURCE_TABLE as (
select '003Preliminary Examination Plan' as numberString
union all select 'Coordination005' as numberString
union all select 'Balance1000sheet' as numberString
union all select '1300 456 678' as numberString
union all select '(012) 995 8322 ' as numberString
union all select '073263 6122,' as numberString
),
FIRST_CHAR_PROCESSED as (
select
len(numberString) as currentStringLength,
isNull(cast(try_cast(replace(left(numberString, 1),' ','z') as tinyint) as nvarchar),'') as firstCharAsNumeric,
cast(isNull(cast(try_cast(nullIf(left(numberString, 1),'') as tinyint) as nvarchar),'') as nvarchar(4000)) as newString,
cast(substring(numberString,2,len(numberString)) as nvarchar) as remainingString
from SOURCE_TABLE
union all
select
len(remainingString) as currentStringLength,
cast(try_cast(replace(left(remainingString, 1),' ','z') as tinyint) as nvarchar) as firstCharAsNumeric,
cast(isNull(newString,'') as nvarchar(3999)) + isNull(cast(try_cast(nullIf(left(remainingString, 1),'') as tinyint) as nvarchar(1)),'') as newString,
substring(remainingString,2,len(remainingString)) as remainingString
from FIRST_CHAR_PROCESSED fcp2
where fcp2.currentStringLength > 1
)
select
newString
,* -- comment this out when required
from FIRST_CHAR_PROCESSED
where currentStringLength = 1
So what's going on here?
Basically in our CTE we are selecting the first character and using try_cast (see docs) to cast it to a tinyint (which is a large enough data type for a single-digit numeral). Note that the type-casting rules in SQL Server say that an empty string (or a space, for that matter) will resolve to zero, so the nullif is added to force spaces and empty strings to resolve to null (see discussion) (otherwise our result would include a zero character any time a space is encountered in the source data).
The CTE also returns everything after the first character - and that becomes the input to our recursive call on the CTE; in other words: now let's process the next character.
Lastly, the field newString in the CTE is generated (in the second SELECT) via concatenation. With recursive CTEs the data type must match between the two SELECT statements for any given column - including the column size. Because we know we are adding (at most) a single character, we are casting that character to nvarchar(1) and we are casting the newString (so far) as nvarchar(3999). Concatenated, the result will be nvarchar(4000) - which matches the type casting we carry out in the first SELECT.
If you run this query and exclude the WHERE clause, you'll get a sense of what's going on - but the rows may be in a strange order. (You won't necessarily see all rows relating to a single input value grouped together - but you should still be able to follow).
Hope it's an interesting option that may help a few people wanting a strictly expression-based solution.
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow