How "SQL Server" finds numbers in a string

Can be achieved by writing a custom function, the following provides two ways to solve:

1, through regular matching, find the number in the string, one by one to spell up

/*method One: one to find out*/CREATE FUNCTION [dbo].[Fun_getnumpart] ( @Str NVARCHAR(MAX) )RETURNS NVARCHAR(MAX) as    BEGIN        DECLARE @Start INT; DECLARE @End INT; DECLARE @Part NVARCHAR(MAX)        SET @Start = PATINDEX('%[0-9]%',@Str); SET @End = PATINDEX('%[0-9]%',SUBSTRING(@Str,@Start+1,LEN(@Str)- @Start)); SET @Part = SUBSTRING(@Str,@Start,1)          while  @End >0         BEGIN                 SET @Start = @start+@End                SET @Part = @Part+SUBSTRING(@Str,@Start,1)                       SET @End = PATINDEX('%[0-9]%',SUBSTRING(@Str,@Start+1,LEN(@Str)- @Start)); END                    RETURN  @Part; END;

2, through the regular reverse matching, find a string of non-numbers, one to remove

/*method Two: one to remove*/CREATE FUNCTION [dbo].[Fun_getnumpart]  ( @Str NVARCHAR(MAX) ) RETURNS NVARCHAR(MAX)    as   BEGIN        while PATINDEX('%[^0-9]%',@Str)>0       BEGIN           SET @Str=STUFF(@Str,PATINDEX('%[^0-9]%',@Str),1,"')--remove non-numeric characters    END       RETURN @Str END   

Effects such as:

Extended:

Can be replaced by the function of '%[0-9]% ' to the '%[a-z]% ', '%[-do]% ', change to find letters, find Chinese function

How "SQL Server" finds numbers in a string