SQL对数字和字符串排序

Question

Hi I have interesting problem, I have about 1500 records within a table. the format of column I need to sort against is

String Number.number.(optional number) (optional string)

In reality this could look like this:

AB 2.10.19
AB 2.10.2
AB 2.10.20 (I)
ACA 1.1
ACA 1.9 (a) V

I need a way to sort these so that instead of

AB 2.10.19
AB 2.10.2
AB 2.10.20 (I)

I get this

AB 2.10.2
AB 2.10.19
AB 2.10.20 (I)

Because of the lack of standard formatting I'm at a loss as to how I can sort this via SQL.

I'm at the point of just manually identifying a new int column to denote the sorting value, unless anyone has any suggestion?

I'm using SQL Server 2008 R2

Answer 1

Try this:

SELECT   column
FROM     table
ORDER BY CASE WHEN SUBSTRING(column,LEN(column)-1,1) = '.'
              THEN 0
              ELSE 1
              END, column

This will put any strings that have a . in the second to last position first in the ordering.

Edit:

On second thought, this won't work with the leading 'AB', 'ACA' etc. Try this instead:

SELECT   column
FROM     table
ORDER BY SUBSTRING(column,1,2), --This will deal with leading letters up to 2 chars
         CASE WHEN SUBSTRING(column,LEN(column)-1,1) = '.'
              THEN 0
              ELSE 1
              END, 
         Column

Edit2:

To also compensate for the second numeric set, use this:

SELECT   column
FROM     table
ORDER BY substring(column,1,2),
CASE WHEN substring(column,charindex('.',column) + 2,1) = '.' and substring(column,len(column)-1,1) = '.' THEN 0 
     WHEN substring(column,charindex('.',column) + 2,1) = '.' and substring(column,len(column)-1,1) <> '.' THEN 1
     WHEN substring(column,charindex('.',column) + 2,1) <> '.' and substring(column,len(column)-1,1) = '.' THEN 2
ELSE 3 END, column

Basically, this is a manual way to force hierarchical ordering by accounting for each condition.

Answer 2

You would need to sort on the first text token, then on the second text token (which is not a number, its a string comprising some numbers) then optionally on any remaining text.

To make the 2nd token sort correctly (like a version number I presume) you can use a hierarchyid:

with t(f) as (
    select 'AB 2.10.19' union all
    select 'AB 2.10.2' union all
    select 'AB 2.10.20 (I)' union all
    select 'AB 2.10.20 (a) Z' union all
    select 'AB 2.10.21 (a)' union all
    select 'ACA 1.1' union all
    select 'ACA 1.9 (a) V' union all
    select 'AB 4.1'
)

select * from t
order by
    left(f, charindex(' ', f) - 1),
    cast('/' + replace(substring(f, charindex(' ', f) + 1, patindex('%[0-9] %', f + ' ') - charindex(' ', f)) , '.', '/') + '/' as hierarchyid),
    substring(f, patindex('%[0-9] %', f + ' ') + 1, len(f))

---

f
----------------
AB 2.10.2
AB 2.10.19
AB 2.10.20 (a) Z
AB 2.10.20 (I)
AB 2.10.21 (a)
AB 4.1
ACA 1.1
ACA 1.9 (a) V

Answer 3

add text for the same length

SELECT   column
FROM     table
ORDER BY left(column + replicate('*', 100500), 100500)

Answer 4

--get the start and end position of numeric in the string
with numformat as 
(select val,patindex('%[0-9]%',val) strtnum,len(val)-patindex('%[0-9]%',reverse(val))+1 endnum 
 from t
 where patindex('%[0-9]%',val) > 0) --where condition added to exclude records with no numeric part in them
--get the substring based on the previously calculated start and end positions
,substrng_to_sort_on as 
(select val, substring(val,strtnum,endnum-strtnum+1) as sub from numformat)
--Final query to sort based on the 1st,2nd and the optional 3rd numbers in the string
select val 
from substrng_to_sort_on
order by 
cast(substring(sub,1,charindex('.',sub)-1) as numeric), --1st number in the string
cast(substring(sub,charindex('.',sub)+1,charindex('.',reverse(sub))) as numeric), --second number in the string
cast(reverse(substring(reverse(sub),1,charindex('.',reverse(sub))-1)) as numeric) --third number in the string

Sample demo