我有一个函数lpad有x个小数 创建函数[dbo].[LPAD_DEC] （ ——在这里添加函数的参数 @pad nvarchar (MAX), @string nvarchar (MAX), @length int, @dec int ） 返回nvarchar (max) 作为 开始 ——在这里声明返回变量 声明@resp nvarchar(max)

IF LEN(@string)=@length
BEGIN
    IF CHARINDEX('.',@string)>0
    BEGIN
        SELECT @resp = CASE SIGN(@string)
            WHEN -1 THEN
                -- Nros negativos grandes con decimales
                concat('-',SUBSTRING(replicate(@pad,@length),1,@length-len(@string)),ltrim(str(abs(@string),@length,@dec)))
            ELSE
                -- Nros positivos grandes con decimales
                concat(SUBSTRING(replicate(@pad,@length),1,@length-len(@string)),ltrim(str(@string,@length,@dec)))                  
            END
    END
    ELSE
    BEGIN
        SELECT @resp = CASE SIGN(@string)
            WHEN -1 THEN
                --Nros negativo grande sin decimales
                concat('-',SUBSTRING(replicate(@pad,@length),1,(@length-3)-len(@string)),ltrim(str(abs(@string),@length,@dec)))
            ELSE
                -- Nros positivos grandes con decimales
                concat(SUBSTRING(replicate(@pad,@length),1,@length-len(@string)),ltrim(str(@string,@length,@dec)))                  
            END                     
    END
END
ELSE
    IF CHARINDEX('.',@string)>0
    BEGIN
        SELECT @resp =CASE SIGN(@string)
            WHEN -1 THEN
                -- Nros negativos con decimales
                concat('-',SUBSTRING(replicate(@pad,@length),1,@length-len(@string)),ltrim(str(abs(@string),@length,@dec)))
            ELSE
                --Ntos positivos con decimales
                concat(SUBSTRING(replicate(@pad,@length),1,@length-len(@string)),ltrim(str(abs(@string),@length,@dec))) 
            END
    END
    ELSE
    BEGIN
        SELECT @resp = CASE SIGN(@string)
            WHEN -1 THEN
                -- Nros Negativos sin decimales
                concat('-',SUBSTRING(replicate(@pad,@length-3),1,(@length-3)-len(@string)),ltrim(str(abs(@string),@length,@dec)))
            ELSE
                -- Nros Positivos sin decimales
                concat(SUBSTRING(replicate(@pad,@length),1,(@length-3)-len(@string)),ltrim(str(abs(@string),@length,@dec)))
            END
    END
RETURN @resp

END

这是我通常如何填充一个varchar

WHILE Len(@String) < 8
BEGIN
    SELECT @String = '0' + @String
END

我用这个。它允许您确定想要的结果长度，以及如果没有提供默认填充字符。当然，您可以为遇到的任何最大值定制输入和输出的长度。

/*===============================================================
 Author         : Joey Morgan
 Create date    : November 1, 2012
 Description    : Pads the string @MyStr with the character in 
                : @PadChar so all results have the same length
 ================================================================*/
 CREATE FUNCTION [dbo].[svfn_AMS_PAD_STRING]
        (
         @MyStr VARCHAR(25),
         @LENGTH INT,
         @PadChar CHAR(1) = NULL
        )
RETURNS VARCHAR(25)
 AS 
      BEGIN
        SET @PadChar = ISNULL(@PadChar, '0');
        DECLARE @Result VARCHAR(25);
        SELECT
            @Result = RIGHT(SUBSTRING(REPLICATE('0', @LENGTH), 1,
                                      (@LENGTH + 1) - LEN(RTRIM(@MyStr)))
                            + RTRIM(@MyStr), @LENGTH)

        RETURN @Result

      END

你的里程可能会有所不同。: -) 乔伊摩根 一级程序设计/分析主任 WellPoint医疗补助事业单位

select right(replicate(@padchar, @len) + @str, @len)

要提供四舍五入到小数点后两位，但如果需要，右补零，我有:

DECLARE @value = 20.1
SET @value = ROUND(@value,2) * 100
PRINT LEFT(CAST(@value AS VARCHAR(20)), LEN(@value)-2) + '.' + RIGHT(CAST(@value AS VARCHAR(20)),2)

如果有人能想出一种更简洁的方法，那将是非常感激的——上面的方法似乎很笨拙。

注意:在本例中，我使用SQL Server以HTML格式发送电子邮件报告，因此希望格式化信息，而不涉及额外的工具来解析数据。

下面是我的解决方案，它避免了截断字符串并使用普通的SQL。感谢@AlexCuse， @Kevin和@Sklivvz，他们的解决方案是这段代码的基础。

 --[@charToPadStringWith] is the character you want to pad the string with.
declare @charToPadStringWith char(1) = 'X';

-- Generate a table of values to test with.
declare @stringValues table (RowId int IDENTITY(1,1) NOT NULL PRIMARY KEY, StringValue varchar(max) NULL);
insert into @stringValues (StringValue) values (null), (''), ('_'), ('A'), ('ABCDE'), ('1234567890');

-- Generate a table to store testing results in.
declare @testingResults table (RowId int IDENTITY(1,1) NOT NULL PRIMARY KEY, StringValue varchar(max) NULL, PaddedStringValue varchar(max) NULL);

-- Get the length of the longest string, then pad all strings based on that length.
declare @maxLengthOfPaddedString int = (select MAX(LEN(StringValue)) from @stringValues);
declare @longestStringValue varchar(max) = (select top(1) StringValue from @stringValues where LEN(StringValue) = @maxLengthOfPaddedString);
select [@longestStringValue]=@longestStringValue, [@maxLengthOfPaddedString]=@maxLengthOfPaddedString;

-- Loop through each of the test string values, apply padding to it, and store the results in [@testingResults].
while (1=1)
begin
    declare
        @stringValueRowId int,
        @stringValue varchar(max);

    -- Get the next row in the [@stringLengths] table.
    select top(1) @stringValueRowId = RowId, @stringValue = StringValue
    from @stringValues 
    where RowId > isnull(@stringValueRowId, 0) 
    order by RowId;

    if (@@ROWCOUNT = 0) 
        break;

    -- Here is where the padding magic happens.
    declare @paddedStringValue varchar(max) = RIGHT(REPLICATE(@charToPadStringWith, @maxLengthOfPaddedString) + @stringValue, @maxLengthOfPaddedString);

    -- Added to the list of results.
    insert into @testingResults (StringValue, PaddedStringValue) values (@stringValue, @paddedStringValue);
end

-- Get all of the testing results.
select * from @testingResults;