Question

Hy，请帮忙，我有12个项目的完整序列，我通过2个不同的组（12345和54321）识别这些项目。现在，我需要确定项目“12345”的第一个序列在4处停止并在10处重新开始。这样的事情：

我有这张桌子：

------------------
|seq  |partNumber|
------------------
| 1   |  12345   |
| 2   |  12345   |
| 3   |  12345   |
| 4   |  12345   |
| 10  |  12345   |
| 11  |  12345   |
| 12  |  12345   |
| 5   |  54321   |
| 6   |  54321   |
| 7   |  54321   |
| 8   |  54321   |
| 9   |  54321   |
------------------

我需要找到这个结果：

------------
|Start|Stop|
------------
|  5  |  9 | (partnumber:12345)
------------

我使用的查询：

select start, stop from (
 select m.partNumber + 1 as start,
    (select min(partNumber) - 1 from seq as x where x.partNumber > m.partNumber) as stop
  from seq as m
    left outer join seq as r on m.partNumber = r.partNumber - 1 where r.partNumber is null) as x
where stop is not null;

但是，这个查询给了我这个结果：

------------
|Start|Stop|
------------
|  9  |  9 | (partnumber:12345)
------------

最终结果：我想确定序列中的“12345”开始于1个结束，4个（休息）重启，10个结束12，我有一个5到9的间隙。“54321”的另一个序列从5个结束开始，9，这里我不要没有差距。

Answer 1

这是结构化查询语言的结构化部分的一个很好的应用程序。

我猜你永远不会有seq小于零或大于仲裁值15624的值。这个猜测很重要：我们需要一个包含该范围内所有基数的表来使缺失数检测起作用。

这是一张表

SELECT A.N + 5*(B.N + 5*(C.N + 5*(D.N + 5*(E.N + 5*(F.N))))) AS seq
  FROM (SELECT 0 AS N UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4) AS A
  JOIN (SELECT 0 AS N UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4) AS B
  JOIN (SELECT 0 AS N UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4) AS C
  JOIN (SELECT 0 AS N UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4) AS D
  JOIN (SELECT 0 AS N UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4) AS E
  JOIN (SELECT 0 AS N UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4) AS F

（如果您使用的是MariaDB，则可以使用Sequence表seq_0_to_15624代替这一部分SQL代码。）

接下来，您需要一种方法来找出每个部件号的seq的最低值和最高值。你是这样做的。

       SELECT partNumber, MIN(seq) minSeq, MAX(seq) maxSeq
         FROM seq
        GROUP BY partNumber

接下来，您需要生成一个表格，显示每个部件编号从最小到最大的所有可能序列号：

SELECT cardinals.seq, r.partNumber
  FROM (
        SELECT A.N + 5*(B.N + 5*(C.N + 5*(D.N + 5*(E.N + 5*(F.N))))) AS seq  
          FROM (SELECT 0 AS N UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4) AS A
          JOIN (SELECT 0 AS N UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4) AS B
         JOIN (SELECT 0 AS N UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4) AS C
         JOIN (SELECT 0 AS N UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4) AS D
         JOIN (SELECT 0 AS N UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4) AS E
         JOIN (SELECT 0 AS N UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4) AS F
       ) cardinals
  JOIN (
         SELECT partNumber, MIN(seq) minSeq, MAX(seq) maxSeq
           FROM seq
          GROUP BY partNumber
   ) r ON cardinals.seq >= r.minSeq AND cardinals.seq <= r.maxSeq

最后，您可以将其连接到原始表格，然后WHERE val IS NULL找到您丢失的序列号。

SELECT cardinals.seq, r.partNumber
  FROM (
        SELECT A.N + 5*(B.N + 5*(C.N + 5*(D.N + 5*(E.N + 5*(F.N))))) AS seq  
          FROM (SELECT 0 AS N UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4) AS A
          JOIN (SELECT 0 AS N UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4) AS B
         JOIN (SELECT 0 AS N UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4) AS C
         JOIN (SELECT 0 AS N UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4) AS D
         JOIN (SELECT 0 AS N UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4) AS E
         JOIN (SELECT 0 AS N UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4) AS F
           ) cardinals
  JOIN (
             SELECT partNumber, MIN(seq) minSeq, MAX(seq) maxSeq
               FROM seq
              GROUP BY partNumber
       ) r ON cardinals.seq >= r.minSeq AND cardinals.seq <= r.maxSeq
  LEFT JOIN seq ON cardinals.seq = seq.seq AND r.PartNumber = seq.partNumber
 WHERE seq.seq IS NULL

MYSQL：通过不同的组查找序列中的缺失值

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