Question

我有这样的输入4541或45或124

我想将输入分成这样的数字：

如果输入为1235，我想采用1，2，3，5

我试图在google上搜索，但我找到的所有内容都是将数字从float中分割出来或将字符串更改为数字。

你可以帮忙吗？

Answer 1

方法1：

试试这个

<强>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

使用方法：

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

<强>输出：

1,2,3,4,5

方法2：

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

数字到Word

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

<强>用法：

SELECT dbo.fnIntegerToWords(5) As Word


    Word
    Five

以逗号分隔列表

获取答案

试试这个

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

Answer 2

您需要将其视为字符串。然后你可以使用SUBSTRING获取每个角色。

DECLARE @str VARCHAR(50)
DECLARE @i INT
DECLARE @len INT
DECLARE @char VARCHAR(50)
DECLARE @result VARCHAR(50)

SET @str = '1235'
SET @i = 1
SET @len = LEN(@str)

WHILE @i<= @len
BEGIN
  SET @char = SUBSTRING(@str, @i, 1)
  @result = @char + ", "
  SET @i=@i+1
END

Answer 3

没有必要像其他人那样声明这么多变量，它很简单。这对我有用：试试吧

declare @intvalue int
declare @incremnt int
 declare @data varchar(20)
set @intvalue = 1234
 set @data = CAST(@intvalue as varchar(10))
 set @incremnt = 1
 while(LEN(@data) >= @incremnt)
begin
 select SUBSTRING(@data,@incremnt,1)
set @incremnt = @incremnt + 1
 end

sql将整数拆分为数字

3 个答案: