SQL SERVER – How to Remove All Characters From a String Using T-SQL?

In the recent past I have seen the best interaction and learning happens when we all get into a puzzle and challenge mode. And that just gets the creative juice in our minds and some really innovative solutions are given. So this blog is written like a nice trivia solution and I would love to hear from you on potential alternate solutions to this as you read the simplified solution.

There are many situations where a developer needs to do a lot of string manipulation using T-SQL. There are many things which ideally should be done via SQLCLR.

SQL SERVER - How to Remove All Characters From a String Using T-SQL? chars

One of my blog reader posted a comment recently

I want to get only integer part in string. for example, string contain ab123ce234fe means i want only integer part like 123234. how can i get?

I gave him below suggestion.

SET NOCOUNT ON
DECLARE
@loop INT
DECLARE
@str VARCHAR(8000)
SELECT @str = 'ab123ce234fe'
SET @loop = 0
WHILE @loop < 26
BEGIN
SET
@str = REPLACE(@str, CHAR(65 + @loop), '')
SET @loop = @loop + 1
END
SELECT
@str

Do you think there is a better way to achieve that? Please write it in comments section. It is going to be a great learning opportunity for all.

Reference: Pinal Dave (https://blog.sqlauthority.com)

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15 Comments. Leave new

  • If you expect few instances of characters it may save time to test for characters before going through all 26 letters:

    SET NOCOUNT ON
    DECLARE @loop INT
    DECLARE @str VARCHAR(8000)
    SELECT @str = ‘ab123ce234fe’
    SET @loop = 0
    WHILE @loop < 26 AND PATINDEX('%[a-z]%',@str) 0
    BEGIN
    SET @str = REPLACE(@str, CHAR(65 + @loop), ”)
    SET @loop = @loop + 1
    END
    SELECT @str

    Or even only looping through as many times as non-integer characters you have:

    SET NOCOUNT ON
    DECLARE @str VARCHAR(8000) = ‘ab123ce234fe’
    DECLARE @pat INT = PATINDEX(‘%[a-z]%’,@str)
    WHILE @pat 0
    BEGIN
    SET @str = LEFT(@str,@pat-1) + RIGHT(@str,LEN(@str)-@pat)

    SET @pat = PATINDEX(‘%[a-z]%’,@str)
    END
    SELECT @str

    Reply
  • This can also be done by using below query

    CREATE FUNCTION udf_extractInteger(@string VARCHAR(2000))
    RETURNS VARCHAR(2000)
    AS

    BEGIN
    DECLARE @count int
    DECLARE @intNumbers VARCHAR(1000)
    SET @count = 0
    SET @intNumbers = ”

    WHILE @count =’0′ and SUBSTRING (@string, @count, 1) <='9'
    BEGIN
    SET @intNumbers = @intNumbers + SUBSTRING (@string, @count, 1)
    END
    SET @count = @count + 1
    END
    RETURN @intNumbers
    END
    GO

    select dbo.udf_extractInteger('ab123ce234fe')

    Reply
  • DECLARE @KeepValues AS VARCHAR(50),
    @Str AS VARCHAR(500)
    SET @KeepValues = ‘%[^a-z]%’
    WHILE PATINDEX(@KeepValues, @Str) > 0
    SET @Str = STUFF(@Str, PATINDEX(@KeepValues, @Str), 1, ”)

    Reply
    • short and simple solution !! Thanks for sharing :)

      Reply
      • For this case, its going to be as following:
        DECLARE @KeepValues VARCHAR(50),@Str VARCHAR(500) = ‘ab123ce234fe’
        SET @KeepValues = ‘%[^a-z]%’ — if you want only characters in the output
        SET @KeepValues = ‘%[^0-9]%’ — If you want only integers in the output
        WHILE PATINDEX(@KeepValues, @Str) > 0
        SET @Str = STUFF(@Str, PATINDEX(@KeepValues, @Str), 1, ”)
        select @Str

      • and suppose string contains ‘ab123cE234Fe’ then how should we handle it?

    • Stephen Yeadon
      August 4, 2015 4:12 pm

      nice solution

      Reply
  • SET NOCOUNT ON
    DECLARE @loop INT
    DECLARE @str VARCHAR(8000)
    DECLARE @strs VARCHAR(8000)
    SELECT @str = ‘ab123ce234fe’
    SET @loop = 0
    set @strs=”
    WHILE @loop < LEN(@str)
    BEGIN
    if(ASCII(SUBSTRING(@str,@loop+1,1)) between 48 and 57)
    begin
    set @strs=@strs+SUBSTRING(@str,@loop+1,1)
    end
    SET @loop = @loop + 1
    END
    print @strs

    Reply
  • Randeep Singh Matharu
    August 4, 2015 4:24 pm

    Below is another Solution, not neccessarily easier to write or especially good but great to see another alternative:

    Declare @strtest varchar(8000)
    Declare @StrFinal varchar(8000)
    Declare @start int

    select @strtest = ‘ab123ce234fe’
    set @StrFinal = ”
    set @start = 1

    while @start < 13

    if isnumeric(SUBSTRING(@strtest,@start,1))= 1

    Begin

    set @StrFinal = @StrFinal + SUBSTRING(@strtest,@start,1)
    set @start = @start + 1

    End

    else set @start = @start + 1

    select @StrFinal

    Reply
  • Scott Williams
    August 5, 2015 3:45 am

    CREATE Function [fnRemoveNonNumericCharacters](@strText VARCHAR(1000))
    RETURNS VARCHAR(1000)
    AS
    BEGIN
    WHILE PATINDEX(‘%[^0-9]%’, @strText) > 0
    BEGIN
    SET @strText = STUFF(@strText, PATINDEX(‘%[^0-9]%’, @strText), 1, ”)
    END
    RETURN @strText
    END

    Reply
  • Chris Munsell
    August 5, 2015 8:36 pm

    Mine is pretty close to Scott Williams… but there is a little difference.
    CREATE Function [dbo].[RemoveAlphaCharacters](@Temp VarChar(5000))
    Returns VarChar(1000)
    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

    Reply
  • If we had auxiliary table like, let’s say, numbers which contains numbers from 1 to
    then we would use the next (just for variety’s sake :)

    declare @str varchar (100) = ‘se345dsfg43dg435dfgh’;

    select ” + symbol
    from (
    select number, SUBSTRING (@str, number, 1) as [symbol]
    from dbo.numbers
    where number <= LEN (@str)
    and SUBSTRING (@str, number, 1) like '[0-9]'
    )
    data
    order by number
    for xml path ('')

    Reply
  • i have somethink like
    1)100w-123 i need output like 100-123
    2)1110A123w i need output like 11110123

    the above queries work for 2) case
    can anyone help on 1 st case

    Reply
    • Use the below code

      SET NOCOUNT ON
      DECLARE @loop INT
      DECLARE @str VARCHAR(8000)
      SELECT @str = ‘ab123ce234fe’
      SET @loop = 0
      WHILE @loop < 26 AND PATINDEX('%[a-z]%',@str) 0
      BEGIN
      SET @str = REPLACE(@str, CHAR(65 + @loop), ”)
      SET @loop = @loop + 1
      END
      SELECT @str

      Reply
  • Hello, I need help for solving the following problem: What is the appropriate way to replace:
    1 / 2 / 3 / 4 / 5 / 6 / 7 / 8 / 9 / 10 / 11 / 12 / 13 / 14 / 15 D / 16 D / 17 D / 18 D with
    text 1,text 2,text 3,text 4,text 5,text 6,text 7,text 8,text 9,text 10,text 11,text 12, text 13, another text 15,another text 16, another text 17, another text 18

    Thank you.

    Reply

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