Tsql 按顺序排列行并填补缺少行的空白

我遇到了一个表格中缺少行的问题，这让我很头疼

作为基础数据，我有下表：

declare @table table
(
  id1   int,
  id2   int,
  ch    char(1) not null,
  val   int     
)

insert into @table values (1112, 121, 'A', 12)
insert into @table values (1351, 121, 'A', 13)
insert into @table values (1411, 121, 'B', 81)
insert into @table values (1312, 7, 'C', 107)
insert into @table values (1401, 2, 'A', 107)
insert into @table values (1454, 2, 'D', 107)
insert into @table values (1257, 6, 'A', 1)
insert into @table values (1269, 6, 'B', 12)
insert into @table values (1335, 6, 'C', 12)
insert into @table values (1341, 6, 'D', 5)
insert into @table values (1380, 6, 'A', 3)

输出应按id2排序，并遵循固定的ch序列，该序列应重复，直到下一个id2开始

顺序：

'A'
'B'
'C'
'D'

如果序列或模式被中断，它应该用null填充缺少的行，这样我就可以得到这个结果表：

id1     id2     ch     val
----------------------------    
1112    121     'A'    12
NULL    121     'B'    NULL
NULL    121     'C'    NULL
NULL    121     'D'    NULL
1351    121     'A'    13
1411    121     'B'    81
NULL    121     'C'    NULL
NULL    121     'D'    NULL
NULL    7       'A'    NULL
NULL    7       'B'    NULL
1312    7       'C'    107
NULL    7       'D'    NULL
1401    2       'A'    107
NULL    2       'B'    NULL
NULL    2       'C'    NULL
1454    2       'D'    107

等等

我要寻找的是一种不需要迭代就可以做到这一点的方法

我希望有人能帮忙

---

提前谢谢

我对您的输出有点困惑，请尝试以下方法：

更新

   DECLARE @table TABLE
    (
      row INT IDENTITY(1, 1) ,
      id1 INT ,
      id2 INT ,
      ch CHAR(1) NOT NULL ,
      val INT
    );

 DECLARE @Sequence TABLE ( ch3 CHAR(1) NOT NULL );


 INSERT INTO @Sequence
 VALUES ( 'A' );
 INSERT INTO @Sequence
 VALUES ( 'B' );
 INSERT INTO @Sequence
 VALUES ( 'C' );
 INSERT INTO @Sequence
 VALUES ( 'D' );

 INSERT INTO @table
 VALUES ( 1112, 121, 'A', 12 );
 INSERT INTO @table
 VALUES ( 1351, 121, 'A', 13 );
 INSERT INTO @table
 VALUES ( 1411, 121, 'B', 81 );
 INSERT INTO @table
 VALUES ( 1312, 7, 'C', 107 );
 INSERT INTO @table
 VALUES ( 1401, 2, 'A', 107 );
 INSERT INTO @table
 VALUES ( 1454, 2, 'D', 107 );
 INSERT INTO @table
 VALUES ( 1257, 6, 'A', 1 );
 INSERT INTO @table
 VALUES ( 1269, 6, 'B', 12 );
 INSERT INTO @table
 VALUES ( 1335, 6, 'C', 12 );
 INSERT INTO @table
 VALUES ( 1341, 6, 'D', 5 );
 INSERT INTO @table
 VALUES ( 1380, 6, 'A', 3 );




 SELECT r.id1 ,
        fin.id2 ,
        ch3 ,
        r.val
 FROM   ( SELECT    *
          FROM      ( SELECT    CASE WHEN r.chd - l.chd = 1 THEN 0
                                     ELSE 1
                                END [gap in sq] ,
                                l.*
                      FROM      ( SELECT    id2 ,
                                            ASCII(ch) chd ,
                                            ch ,
                                            val ,
                                            id1 ,
                                            row
                                  FROM      @table
                                ) AS l
                                LEFT JOIN ( SELECT  id2 ,
                                                    ASCII(ch) chd ,
                                                    row
                                            FROM    @table
                                          ) AS r ON l.row = r.row - 1
                    ) AS temp ,
                    @Sequence s
          WHERE     temp.[gap in sq] = 1
                    OR ( temp.[gap in sq] = 0
                         AND s.ch3 = temp.ch
                       )
        ) AS fin
        LEFT JOIN @table r ON r.id2 = fin.id2
                              AND r.id1 = fin.id1
                              AND r.ch = fin.ch3

---

如果您能够在表中添加一个额外的列，该列定义了哪些[id2]是同一序列的一部分，您可以尝试以下操作：

declare @table table
(
  id1   int,
  id2   int,
  ch    char(1) not null,
  val   int,
  category  int -- extra column
)

insert into @table values (1112, 121, 'A', 12, 1)
insert into @table values (1351, 121, 'A', 13, 2)
insert into @table values (1411, 121, 'B', 81, 2)
insert into @table values (1312, 7, 'C', 107, 3)
insert into @table values (1401, 2, 'A', 107, 4)
insert into @table values (1454, 2, 'D', 107, 4)
insert into @table values (1257, 6, 'A', 1, 5)
insert into @table values (1269, 6, 'B', 12, 5)
insert into @table values (1335, 6, 'C', 12, 5)
insert into @table values (1341, 6, 'D', 5, 5)
insert into @table values (1380, 6, 'A', 3, 5)


DECLARE @sequence table (seq varchar(1))
INSERT INTO @sequence values ('A'), ('B'), ('C'), ('D')


SELECT b.id1, a.id2, a.seq, b.val, a.category
INTO #T1
FROM (
    SELECT *
    FROM  @table
    CROSS JOIN @sequence
) A
LEFT JOIN (
    SELECT * FROM @table
) B
    ON 1=1
    AND a.id1 = b.id1
    AND a.id2 = b.id2
    AND a.seq = b.ch
    AND a.val = b.val


;WITH rem_duplicates AS (
    SELECT *, dup = ROW_NUMBER() OVER (PARTITION by id2, seq, category ORDER BY id1 DESC)
    FROM #T1
) DELETE FROM rem_duplicates WHERE dup > 1


SELECT * FROM #T1 ORDER BY id2 DESC, category ASC, seq ASC

DROP TABLE #T1

---

解决方案可能是：

declare @table table (  id1   int,  id2   int,  ch    char(1) not null,  val   int     )
insert into @table values (1112, 121, 'A', 12)
    ,(1351, 121, 'A', 13),(1411, 121, 'B', 81),(1312, 7, 'C', 107),(1401, 2, 'A', 107)
    ,(1454, 2, 'D', 107),(1257, 6, 'A', 1),(1269, 6, 'B', 12),(1335, 6, 'C', 12)
    ,(1341, 6, 'D', 5),(1380, 6, 'A', 3)
;with foo as
(select 
    *
    ,row_number() over (partition by id2 order by id1) rwn
    ,ascii(isnull(lag(ch,1) over (partition by id2 order by id1),'A'))-ascii('A')  prev
    ,count(*) over (partition by id2,ch) nr
    ,ascii(ch)-ascii('A') cur
from @table

)
,bar as
(
select 
*,case when cur<=prev and rwn>1 then 4 else 0 end + cur-prev step
from foo
)
,foobar as
(
select *,sum(step) over (partition by id2 order by id1 rows unbounded preceding) rownum
from bar
)
,iterations as
(
select id2,max(nr) nr from foo
group by id2
)
,blanks as
(
select
id2,ch chnr,char(ch+ascii('A') )ch,ROW_NUMBER() over (partition by id2 order by c.nr,ch)-1 rownum,c.nr
from iterations a
inner join (values (1),(2),(3),(4),(5),(6),(7),(8),(9),(10)) c(nr)
on c.nr<=a.nr
cross join (values (0),(1),(2),(3)) b(ch)
)
select
b.id1,a.id2,a.ch,b.val
from blanks a
left join foobar b
on a.id2=b.id2 and a.rownum=b.rownum 
order by a.id2,a.rownum

我首先进行查询foo，它查看行号并获取每个id2的ch的前一个值

然后，bar将查找行之间缺少的值的数量。例如，如果前一个是A，当前是c，则有2个。如果前一个是A，当前是A，则有4个

foobar然后添加步骤，从而对原始行进行编号，这些行应该位于最终输出中

迭代次数统计ABCD行应出现的次数

空白然后是所有最终行，即对于每个id2，它输出应该在最终输出中的所有ABCD行，并在rownum中对它们进行编号

---

最后，我在id2和rownum上用空格离开了join foobar。这样我们就得到了正确的行数，并且输出了原始数据中有值的位置。

谢谢，非常感谢！但是1351和1411是在相同的序列中。所以1411应该紧跟在1351之后的第6行，而不是第10行谢谢你的输入，Valerica！。我想，额外列的定义是我无法绕过迭代的地方。非常感谢您的解决方案和解释！你真是个天才！