复杂查询

Question

我有一个表：

Number | Name 
__________________ 

100  | A 
101  | B 
102  | C 
103  | D 
104  | E 
..................  
105  | F 
106  | G 
108  | I 
109  | J 
..................  
110  | K 
111  | L 
114  | O 
.................. 
115  | P 
116  | Q 
...  | ... 

有Number 0。4和其他字符是相同的是在一组，而那些有5间最后一个字符的所有行.. 9在另一个组中。这类团体在此表：

Group 100 
    Group 105 
    Group 110 
    Group 115 
    Group ... 

我怎样才能找到不全组（缺少的成员）？

对于以上（显示为一个例子）表，结果是：

Group 105 [missing members: 107] 
    Group 110 [missing members: 112,113] 

Answer 1

这里是你如何能做到这

-- sample data 
; with tbl as 
(
    select Number = 100 union all 
    select Number = 101 union all 
    select Number = 102 union all 
    select Number = 103 union all 
    select Number = 104 union all 
    select Number = 105 union all 
    select Number = 106 union all 
    select Number = 108 union all 
    select Number = 109 union all 
    select Number = 110 union all 
    select Number = 111 
), 
-- number tables (here you only required 0 to 4 as you are grouping by 5) 
num as 
(
    select n = 0 union all 
    select n = 1 union all 
    select n = 2 union all 
    select n = 3 union all 
    select n = 4 
) 
-- the query 
select [Group] = t.Number/5 * 5, misisng = max(stuff(m.missing, 1, 1, '')) 
from tbl t 
    cross apply 
    (
     select ',' + convert(varchar(10), (t.Number/5 * 5) + n) 
     from num 
     where not exists 
      (
       select * 
       from tbl x 
       where x.Number = (t.Number/5 * 5) + n 
      ) 
     order by n 
     for xml path('')    
    ) m (missing) 
where m.missing is not null 
group by t.Number/5 * 5 

- 如果你想在mssing使用此查询no in separate lines - 查询 选择不同[Group] = t.Number/5 * 5，m.missing from tbl t 交叉应用 （ 选择缺失=（t.Number/5 * 5）+ n的NUM 其中不存在 （ SELECT * 从TBL X 其中x.Number =（t.Number/5 * 5）+ N ） ）米 其中m.missing不为空

Answer 2

它工作正常。

Group 105 [missing members: 107] 
Group 110 [missing members: 112,113] 

我怎么能得到这个结果在单独的记录，平均组110在两个记录。 是这样的：

Group 105 [missing member: 107] 
Group 110 [missing member: 112] 
Group 110 [missing member: 113]