接下来拆分这一串字符串，每个字符插入一个表变量中，最后使用GROUP BY进行分组。
CalNumOfChtInStr
SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO
-- =============================================
-- Author: Insus.NET
-- Create date: 2012-02-23
-- Description: Calculate the number of characters in the string
-- =============================================
CREATE PROCEDURE [dbo].[CalNumOfChtInStr]
(
@Value NVARCHAR(MAX)
)
AS
BEGIN
DECLARE @dum TABLE ([Str] NVARCHAR(2))
DECLARE @I INT = LEN(@Value),@J INT = LEN(@Value)
WHILE @I > 0
BEGIN
--以下函数可参考：http://www.cnblogs.com/insus/archive/2011/06/25/2090231.html
IF [dbo].[IsInteger] (SUBSTRING(@Value, @I, 1)) = 0
BEGIN
RAISERROR('传入字符串包含其它字符，不完全是数字。',16,1)
RETURN
END
SET @I = @I - 1
END
WHILE @J > 0
BEGIN
INSERT INTO @dum VALUES(SUBSTRING(@Value, @J, 1))
SET @J = @J - 1
END
SELECT [Str],COUNT([Str]) AS [Num] FROM @dum GROUP BY [Str]
END
Demo:
EXECUTE [dbo].[CalNumOfChtInStr] '5487554127489423454'
结果：
以下文字更新于2012-02-24 09:40
分析以上的存储过程，考虑到性能问题，它在判断是否包含有其它字符时，循环一次字符串，然后又循环一次将每一个字符插入表变量中。以下修改正此点只做循环一次。
CalNumOfChtInStr
SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO
-- =============================================
-- Author: Insus.NET
-- Create date: 2012-02-23
-- Update date: 2012-02-24
-- Description: Calculate the number of characters in the string
-- =============================================
ALTER PROCEDURE [dbo].[CalNumOfChtInStr]
(
@Value NVARCHAR(MAX)
)
AS
BEGIN
DECLARE @dum TABLE ([Str] NVARCHAR(2))
DECLARE @I INT = LEN(@Value)
WHILE @I > 0
BEGIN
INSERT INTO @dum VALUES(SUBSTRING(@Value, @I, 1))
SET @I = @I - 1
END
--以下函数可参考：http://www.cnblogs.com/insus/archive/2011/06/25/2090231.html
IF EXISTS(SELECT TOP 1 1 FROM @dum WHERE [dbo].[IsInteger]([Str]) = 0)
BEGIN
RAISERROR('传入字符串包含其它字符，不完全是数字。',16,1)
RETURN
END
SELECT [Str],COUNT([Str]) AS [Num] FROM @dum GROUP BY [Str]
END
以下内容于2012-04-29 10:44分添加：
如果想参考C#版本
https://www.jb51.net/article/30211.htm 
