2

我有一张桌子，可以保存我们所有商店的所有订单。我写了一个查询来检查每个商店的顺序订单。看起来是这样的。

select WebStoreID, min(webordernumber), max(webordernumber), count(webordernumber) 
from orders
where ordertype = 'WEB' 
group by WebStoreID

我可以通过此查询检查所有订单是否存在。网络订单号是 1...n 中的数字。

如何在不加入临时/不同表的情况下编写查询以查找丢失的订单？

4

4 回答 4

6

您可以自行加入表以检测没有前一行的行：

select cur.*
from orders cur
left join orders prev 
    on cur.webordernumber = prev.webordernumber + 1
    and cur.webstoreid = prev.webstoreid
where cur.webordernumber <> 1
and prev.webordernumer is null

这将检测 1...n 序列中的间隙，但不会检测到重复。

于 2009-04-29T15:22:55.513 回答
4

我会制作一个“从 1 到 n 的所有整数”的辅助表（有关使用 SQL Server 函数制作它的一些方法，请参见http://www.sql-server-helper.com/functions/integer-table.aspx ，但是因为它是你一遍又一遍需要的东西，无论如何我都会把它变成一个真正的表，并且使用任何 SQL 引擎很容易做到这一点，只需一次）然后使用嵌套查询，SELECT value FROM integers WHERE value NOT IN (SELECT webordernumber FROM orders)&c。另请参阅http://www.sqlmag.com/Article/ArticleID/99797/sql_server_99797.html以了解与您类似的问题，“检测数字序列中的间隙”。

于 2009-04-29T15:17:30.830 回答
2

如果你有 rank() 函数但没有 lag() 函数（换句话说，SQL Server），你可以使用它（由http://www.sqlmonster.com/Uwe/Forum.aspx/sql-server建议-programming/10594/Return-gaps-in-a-sequence）：

create table test_gaps_in_sequence (x int)
insert into test_gaps_in_sequence values ( 1 )
insert into test_gaps_in_sequence values ( 2 )
insert into test_gaps_in_sequence values ( 4 )
insert into test_gaps_in_sequence values ( 5 )
insert into test_gaps_in_sequence values ( 8 )
insert into test_gaps_in_sequence values ( 9 )
insert into test_gaps_in_sequence values ( 12)
insert into test_gaps_in_sequence values ( 13)
insert into test_gaps_in_sequence values ( 14)
insert into test_gaps_in_sequence values ( 29)

...

 select lower_bound
         , upper_bound
      from (select upper_bound
                 , rank () over (order by upper_bound) - 1 as upper_rank
              from (SELECT x+n as upper_bound
                      from test_gaps_in_sequence
                         , (SELECT 0 n
                            UNION
                            SELECT -1
                           ) T
                     GROUP BY x+n
                    HAVING MAX(n) = -1
                    ) upper_1
            ) upper_2
         , (select lower_bound
                 , rank () over (order by lower_bound) as lower_rank
              from (SELECT x+n as lower_bound
                      from test_gaps_in_sequence
                         , (SELECT 0 n
                            UNION
                            SELECT 1
                           ) T
                     GROUP BY x+n
                    HAVING MIN(n) = 1
                    ) lower_1
            ) lower_2
      where upper_2.upper_rank = lower_2.lower_rank
      order by lower_bound

...或者，包括“外部限制”：

select lower_bound
     , upper_bound
  from (select upper_bound
             , rank () over (order by upper_bound) - 1 as upper_rank
          from (SELECT x+n as upper_bound
                  from test_gaps_in_sequence
                     , (SELECT 0 n
                        UNION
                        SELECT -1
                       ) T
                 GROUP BY x+n
                HAVING MAX(n) = -1
                ) upper_1
        ) upper_2
   full join (select lower_bound
             , rank () over (order by lower_bound) as lower_rank
          from (SELECT x+n as lower_bound
                  from test_gaps_in_sequence
                     , (SELECT 0 n
                        UNION
                        SELECT 1
                       ) T
                 GROUP BY x+n
                HAVING MIN(n) = 1
                ) lower_1
        ) lower_2
   on upper_2.upper_rank  = lower_2.lower_rank
     order by coalesce (lower_bound, upper_bound)
于 2011-02-19T13:00:09.247 回答
1

如果您的数据库支持分析函数，那么您可以使用如下查询：

select prev+1, curr-1 from
( select webordernumber curr,
         coalesce (lag(webordernumber) over (order by webordernumber), 0) prev
  from   orders
)
where prev != curr-1;

输出将显示差距，例如

prev+1 curr-1
------ ------
     3      7

意味着缺少 3 到 7 的数字。

于 2009-04-29T15:17:48.987 回答