使用sql拆分alpha和数字

时间:2014-11-19 06:21:33

标签: sql-server sql-server-2005 split

我有一张桌子,它有3列。 first column 是包含值(数字)和单位(百分比等等)的数据,第二列是 numeric column ,第三是 Unit column 。我想要做的是从第一列拆分数字和单位,然后将那些拆分数据放到指定的列。

这是我的表:

UPDATED Image

我尝试了这个函数:SO link here...,它确实拆分了alpha和numeric,但后来我使用了SQL函数,我的问题是参数必须在字符串STRING中,所以我做的是将其更改为Sub Query但它给了我错误。

示例COde:

SQL功能:

create function [dbo].[GetNumbersFromText](@String varchar(2000))
returns table as return
(
  with C as
  (
    select cast(substring(S.Value, S1.Pos, S2.L) as int) as Number,
           stuff(s.Value, 1, S1.Pos + S2.L, '') as Value
    from (select @String+' ') as S(Value)
      cross apply (select patindex('%[0-9]%', S.Value)) as S1(Pos)
      cross apply (select patindex('%[^0-9]%', stuff(S.Value, 1, S1.Pos, ''))) as S2(L)
    union all
    select cast(substring(S.Value, S1.Pos, S2.L) as int),
           stuff(S.Value, 1, S1.Pos + S2.L, '')
    from C as S
      cross apply (select patindex('%[0-9]%', S.Value)) as S1(Pos)
      cross apply (select patindex('%[^0-9]%', stuff(S.Value, 1, S1.Pos, ''))) as S2(L)
    where patindex('%[0-9]%', S.Value) > 0
  )
  select Number
  from C
)

使用SUB查询选择语句:

declare @S varchar(max)
select number from GetNumbersFromText(Select SomeColm From Table_Name) option (maxrecursion 0)

BTW,我正在使用sql server 2005。

谢谢!

4 个答案:

答案 0 :(得分:10)

如果数字部分始终位于开头,则可以使用:

PATINDEX('%[0-9][^0-9]%', ConcUnit)

获取最后一位数的索引。

因此,这个:

DECLARE @str VARCHAR(MAX) = '4000 ug/ML' 

SELECT LEFT(@str, PATINDEX('%[0-9][^0-9]%', @str )) AS Number,
       LTRIM(RIGHT(@str, LEN(@str) - PATINDEX('%[0-9][^0-9]%', @str ))) As Unit

给你:

Number  Unit
-------------
4000    ug/ML

修改

如果数字数据也包含双精度值,那么您可以使用:

SELECT LEN(@str) - PATINDEX ('%[^0-9][0-9]%', REVERSE(@str))

获取最后一位数的索引

因此,这个:

SELECT LEFT(@str, LEN(@str) - PATINDEX ('%[^0-9][0-9]%', REVERSE(@str)))

为您提供数字部分。

而且:

SELECT LEFT(@str, LEN(@str) - PATINDEX ('%[^0-9][0-9]%', REVERSE(@str))) AS Numeric,
       CASE 
          WHEN CHARINDEX ('%', @str) <> 0 THEN LTRIM(RIGHT(@str, LEN(@str) - CHARINDEX ('%', @str)))
          ELSE LTRIM(RIGHT(@str, PATINDEX ('%[^0-9][0-9]%', REVERSE(@str))))
       END AS Unit

为您提供数字和单位部分。

以下是我对您发布的数据进行的一些测试:

输入:

DECLARE @str VARCHAR(MAX) = '50 000ug/ML'

输出:

Numeric Unit
------------
50 000  ug/ML

输入:

DECLARE @str VARCHAR(MAX) = '99.5%'

输出:

Numeric Unit
------------
99.5    

输入:

DECLARE @str VARCHAR(MAX) = '4000 . 35 % ug/ML'

输出:

Numeric     Unit
------------------
4000 . 35   ug/ML

答案 1 :(得分:3)

这是我的答案。检查SQLFiddle中的输出是否相同。

create TABLE temp
(
      string NVARCHAR(50)
)

INSERT INTO temp (string)
VALUES 
    ('4000 ug\ml'),
    ('2000 ug\ml'),
    ('%'),
    ('ug\ml')

SELECT subsrtunit,LEFT(subsrtnumeric, PATINDEX('%[^0-9]%', subsrtnumeric+'t') - 1)
FROM (
    SELECT subsrtunit = SUBSTRING(string, posofchar, LEN(string)),
  subsrtnumeric = SUBSTRING(string, posofnumber, LEN(string))
    FROM (
        SELECT string, posofchar = PATINDEX('%[^0-9]%', string),
      posofnumber = PATINDEX('%[0-9]%', string)
        FROM temp
    ) d
) t

更新版本以处理 99.5 ug \ ml

create TABLE temp
(
      string NVARCHAR(50)
)

INSERT INTO temp (string)
VALUES 
    ('4000 ug\ml'),
    ('2000 ug\ml'),
    ('%'),
    ('ug\ml'),
    ('99.5 ug\ml')

SELECT subsrtunit,LEFT(subsrtnumeric, PATINDEX('%[^0-9.]%', subsrtnumeric+'t') - 1)
FROM (
    SELECT subsrtunit = SUBSTRING(string, posofchar, LEN(string)),
  subsrtnumeric = SUBSTRING(string, posofnumber, LEN(string))
    FROM (
        SELECT string, posofchar = PATINDEX('%[^0-9.]%', string),
      posofnumber = PATINDEX('%[0-9.]%', string)
        FROM temp
    ) d
) t

更新版本:处理 1 000 ug \ ml,20 000ug \ ml

create TABLE temp
(
      string NVARCHAR(50)
)

INSERT INTO temp (string)
VALUES 
    ('4000 ug\ml'),
    ('2000 ug\ml'),
    ('%'),
    ('ug\ml'),
    ('99.5 ug\ml'),
    ('1 000 ug\ml'),
    ('20 000ug\ml')

SELECT substring(replace(subsrtunit,' ',''),PATINDEX('%[0-9.]%', replace(subsrtunit,' ',''))+1,len(subsrtunit)),
LEFT(replace(subsrtnumeric,' ',''), PATINDEX('%[^0-9.]%', replace(subsrtnumeric,' ','')+'t') - 1)
FROM (
    SELECT subsrtunit = SUBSTRING(string, posofchar, LEN(string)),
  subsrtnumeric = SUBSTRING(string, posofnumber, LEN(string))
    FROM (
        SELECT string, posofchar = PATINDEX('%[^0-9.]%', replace(string,' ','')),
      posofnumber = PATINDEX('%[0-9.]%', replace(string,' ',''))
        FROM temp
    ) d
) t

查看SQLFiddle相同内容。

答案 2 :(得分:1)

这样的事情会起作用吗?根据显示的数据,它看起来会如此。

将其作为选择应用于您的数据集,如果您喜欢结果,则可以从中进行更新。

WITH cte as (SELECT 'ug/mL' ConcUnit, 500 as [Numeric], '' as Unit 
   UNION ALL SELECT '2000 ug/mL',     NULL,             '')

SELECT
    [ConcUnit]                  as [ConcUnit],
    [Numeric]                   as [Original Numeric],
    [Unit]                      as [Original Unit],
    CASE WHEN ConcUnit LIKE '% %' THEN 
        SUBSTRING(ConcUnit, 1, CHARINDEX(' ', ConcUnit) - 1) 
        ELSE [Numeric] END      as [New Numeric],
    CASE WHEN ConcUnit LIKE '% %' 
        THEN SUBSTRING(ConcUnit, CHARINDEX(' ', ConcUnit) + 1, LEN(ConcUnit)) 
        ELSE ConcUnit END       as [New Unit]
FROM cte

答案 3 :(得分:1)

改变@concunit&amp; @unitx分别

 DECLARE @concunit varchar(10)='45.5%'
DECLARE @unitx varchar(10)='%'

BEGIN
SELECT RTRIM(SUBSTRING( @concunit , 1 , CHARINDEX( @unitx , @concunit
                                              ) - 1
                )) AS Number, 
       RTRIM(SUBSTRING( @concunit , CHARINDEX( @unitx , @concunit
                                          ) , LEN( @concunit
                                                 ) - (CHARINDEX( @unitx , @concunit
                                                               ) - 1)
                )) AS Unit

end