sql 将整数拆分为数字

Question

我有这样的输入 4541或 45或 124
我想将输入拆分为其数字，如下所示:

如果输入是 1235 ，我要带1 , 2 , 3 , 5
我试图在谷歌上搜索，但我发现的一切都是将数字从浮点数拆分或将字符串更改为数字。

你能帮忙吗？

Answer 1

方法一:

尝试这个

功能:

IF EXISTS (SELECT * FROM sys.objects WHERE object_id = OBJECT_ID(N'[dbo].[SplitIntoIndividualLetters]') AND type in (N'FN', N'IF', N'TF', N'FS', N'FT')) 
DROP FUNCTION [dbo].[SplitIntoIndividualLetters] 
GO 
CREATE FUNCTION [dbo].[SplitIntoIndividualLetters](@Str varchar(8000)) 
Returns @Lettertable TABLE (letters varchar(8000)) 
AS 
Begin 
;With NumCte As 
( 
Select Number = 1 Union All 
Select Number +1 From NumCte 
Where Number < 1000 
) 
Insert Into @Lettertable(letters) 
Select Substring(@str,Number,1) 
From NumCte 
where Number Between 1 And Len(@str) 
Option (Maxrecursion 0) 
Return 
End 

如何使用:

Declare @str Varchar(50) = '12345' 
DECLARE @listStr VARCHAR(MAX)
SELECT @listStr = COALESCE(@listStr+',' ,'') + letters
FROM dbo.SplitIntoIndividualLetters(@str) 
SELECT @listStr

输出:

1,2,3,4,5

方法二:

我想出了 Donal解决方案:

DECLARE @str VARCHAR(50),@Inc INT,@len INT,@char VARCHAR(50)

SET @str = '12345'
SET @Inc = 1
SET @len = LEN(@str)

WHILE @Inc<= @len
BEGIN
  SET @char = COALESCE(@char+',' ,'') + SUBSTRING(@str, @Inc, 1)  
  SET @Inc=@Inc+1
END

SELECT @char

输出:

1,2,3,4,5

数转字

CREATE FUNCTION fnIntegerToWords(@Number as BIGINT) 
    RETURNS VARCHAR(1024)
AS

BEGIN
      DECLARE @Below20 TABLE (ID int identity(0,1), Word varchar(32))
      DECLARE @Below100 TABLE (ID int identity(2,1), Word varchar(32))
      INSERT @Below20 (Word) VALUES 
                        ( 'Zero'), ('One'),( 'Two' ), ( 'Three'),
                        ( 'Four' ), ( 'Five' ), ( 'Six' ), ( 'Seven' ),
                        ( 'Eight'), ( 'Nine'), ( 'Ten'), ( 'Eleven' ),
                        ( 'Twelve' ), ( 'Thirteen' ), ( 'Fourteen'),
                        ( 'Fifteen' ), ('Sixteen' ), ( 'Seventeen'),
                        ('Eighteen' ), ( 'Nineteen' ) 

       INSERT @Below100 VALUES ('Twenty'), ('Thirty'),('Forty'), ('Fifty'),
                               ('Sixty'), ('Seventy'), ('Eighty'), ('Ninety')

    declare @belowHundred as varchar(126) 

    if @Number > 99 begin
        select @belowHundred = dbo.fnIntegerToWords( @Number % 100)
    end

    DECLARE @English varchar(1024) = 

    (

      SELECT Case 
        WHEN @Number = 0 THEN  ''

        WHEN @Number BETWEEN 1 AND 19 
          THEN (SELECT Word FROM @Below20 WHERE ID=@Number)

       WHEN @Number BETWEEN 20 AND 99   
         THEN  (SELECT Word FROM @Below100 WHERE ID=@Number/10)+ '-' +
               dbo.fnIntegerToWords( @Number % 10) 

       WHEN @Number BETWEEN 100 AND 999   
         THEN  (dbo.fnIntegerToWords( @Number / 100)) +' Hundred '+
             Case WHEN @belowHundred <> '' THEN 'and ' + @belowHundred else @belowHundred end 

       WHEN @Number BETWEEN 1000 AND 999999   
         THEN  (dbo.fnIntegerToWords( @Number / 1000))+' Thousand '+
             dbo.fnIntegerToWords( @Number % 1000)  

       WHEN @Number BETWEEN 1000000 AND 999999999   
         THEN  (dbo.fnIntegerToWords( @Number / 1000000))+' Million '+
             dbo.fnIntegerToWords( @Number % 1000000) 

       WHEN @Number BETWEEN 1000000000 AND 999999999999   
         THEN  (dbo.fnIntegerToWords( @Number / 1000000000))+' Billion '+
             dbo.fnIntegerToWords( @Number % 1000000000) 

            ELSE ' INVALID INPUT' END
    )

    SELECT @English = RTRIM(@English)

    SELECT @English = RTRIM(LEFT(@English,len(@English)-1))
        WHERE RIGHT(@English,1)='-'

    RETURN (@English)

END 

用法:

SELECT dbo.fnIntegerToWords(5) As Word


    Word
    Five

在逗号分隔列表中获取答案

尝试这个

Declare @str Varchar(50) = '12345' 

DECLARE @listStr VARCHAR(MAX)
SELECT @listStr = COALESCE(@listStr+',' ,'') + dbo.fnIntegerToWords(letters)
FROM dbo.SplitIntoIndividualLetters(@str) 
SELECT @listStr

输出:

One,Two,Three,Four,Five