获取序列中所有缺失的数字

Question

The numbers are originally alpha numeric so I have a query to parse out the numbers:

My query here gives me a list of numbers:

select distinct cast(SUBSTRING(docket,7,999) as INT) from [DHI_IL_Stage].[dbo].[Violation] where InsertDataSourceID='40' and ViolationCounty='Carroll' and SUBSTRING(docket,5,2)='TR' and LEFT(docket,4)='2011' order by 1

Returns the list of numbers parsed out. For example, the number will be 2012TR557. After using the query it will be 557.

I need to write a query that will give back the missing numbers in a sequence.

Answer 1

Here is one approach The following should return one row for each sequence of missing numbers. So, if you series is 3, 5, 6, 9, then it should return:

4     4
7     8

The query is:

with nums as (
       select distinct cast(SUBSTRING(docket, 7, 999) as INT) as n,
              row_number() over (order by cast(SUBSTRING(docket, 7, 999) as INT)) as seqnum
       from [DHI_IL_Stage].[dbo].[Violation]
       where InsertDataSourceID = '40' and
             ViolationCounty = 'Carroll' and
             SUBSTRING(docket,5,2) = 'TR' and
             LEFT(docket, 4) = '2011'
     )
select (nums_prev.n + 1) as first_missing, nums.n - 1 as last_missing
from nums left outer join
     nums nums_prev
     on nums.seqnum = nums_prev.seqnum + 1
where nums.n <> nums_prev.n + 1 ;