查找“滚动”中的最后一个值 与存储过程的顺序?
假设我有一组固定长度的字母字符标识符,例如总是五个字母,并且它们的分配方式是它们总是按顺序递增(GGGGZ --> GGGHA 等)。 现在,如果我到达 ZZZZZ,由于长度是固定的,我必须“翻转”到 AAAAA。 我可能有一个从 ZZZAA 到 AAAAM 的连续块。 我想编写一个存储过程来为我提供“下一个”标识符,在本例中为 AAAAN。
当然,如果我没有这个“滚动”问题,我只需按 DESC 排序并获取最高结果。 但我现在有点不知所措——SQL 不是我最擅长的语言,这一点也于事无补。
如果我有,我可以将其移至我的 C# 调用代码,但存储过程会更合适。
ETA:我想避免更改架构(新列或新表); 我宁愿能够“弄清楚”。 我什至可能更喜欢使用蛮力(例如从最低值开始并递增,直到找到“洞”),尽管这可能会很昂贵。 如果您有一个不修改架构的答案,那么这将是满足我需求的更好解决方案。
Suppose I had a set of alpha-character identifiers of a set length, e.g. always five letters, and they are assigned in such a way that they are always incremented sequentially (GGGGZ --> GGGHA, etc.). Now, if I get to ZZZZZ, since the length is fixed, I must "roll over" to AAAAA. I might have a contiguous block from ZZZAA through AAAAM. I want to write a sproc that will give me the "next" identifier, in this case AAAAN.
If I didn't have this "rolling over" issue, of course, I'd just ORDER BY DESC and grab the top result. But I'm at a bit of a loss now -- and it doesn't help at all that SQL is not my strongest language.
If I have to I can move this to my C# calling code, but a sproc would be a better fit.
ETA: I would like to avoid changing the schema (new column or new table); I'd rather just be able to "figure it out". I might even prefer to do it brute force (e.g. start at the lowest value and increment until I find a "hole"), even though that could get expensive. If you have an answer that does not modify the schema, it'd be a better solution for my needs.
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我认为这里的代码将为您提供下一个值。 我创建了 3 个函数。 该表只是我对 table.column 和 alpha id 的模拟(我使用了 MyTable.AlphaID)。 我假设它正如您所暗示的那样,并且有一个连续的五个字符大写字母字符串块(AlphaID):
顺便说一句,如果您不想假设连续性,您可以按照您的建议进行操作,并且(如果有 ' ZZZZZ' 行)使用序列中的第一个间隙。 将最后一个函数替换为:
Here's code that I think will give you your Next value. I created 3 functions. The table is just my simulation of the table.column with your alpha ids (I used MyTable.AlphaID). I assume that it's as you implied and there is one contiguous block of five-character uppercase alphabetic strings (AlphaID):
By the way, if you don't want to assume contiguity, you can do as you suggested and (if there's a 'ZZZZZ' row) use the first gap in the sequence. Replace the last function with this:
您必须存储序列中最后分配的标识符。
例如,将其存储在另一个具有一列和一列的表中。 一排。
每次分配新标识符时,您都会获取该小表中的值,递增该值,然后将该值存储在主表中,并更新
CurrentMaxId
中的值。通常的注意事项适用于并发、表锁定等。
You'd have to store the last allocated identifier in the sequence.
For example, store it in another table that has one column & one row.
Each time you allocate a new identifier, you'd fetch the value in that tiny table, increment it, and store that value in your main table as well as updating the value in
CurrentMaxId
.The usual caveats apply with respect to concurrency, table-locking, etc.
我想我会尝试将序列存储为整数,然后将其转换为字符串。 或者存储与 alpha 值同时递增的并行整数列。 无论哪种方式,您都可以对整数列进行排序。
I think I'd have tried to store the sequence as an integer, then translate it to string. Or else store a parallel integer column that is incremented at the same time as the alpha value. Either way, you could sort on the integer column.
这里的一个问题是,您无法真正从数据中判断“最后”条目在哪里,除非有更多关于如何删除旧条目的详细信息。
如果我理解正确的话,您将在序列的末尾环绕,这意味着您必须删除一些旧数据以腾出空间。 但是,如果数据没有以完全统一的方式删除,您最终会得到碎片,如下所示:
您会注意到没有明显的下一个值...D 可能是最后创建的值,但也可能也可以是 L 或 S。
最多您可以查找第一个或最后一个缺失元素(使用存储过程执行 x+1 检查,就像在整数序列中查找缺失元素一样),但它不会为滚动列表提供任何特殊结果。
A problem here is that you can't really tell from the data where the "last" entry is unless there is more detail as to how the old entries are deleted.
If I understand correctly, you are wrapping around at the end of the sequence, which means you must be deleting some of your old data to make space. However if the data isn't deleted in a perfectly uniform manner, you'll end up with fragments, like below:
You'll notice that there is no obvious next value...D could be the last value created, but it might also be L or S.
At best you could look for the first or last missing element (use a stored procedure to perform a x+1 check just like you would to find a missing element in an integer sequence), but it's not going to provide any special result for rolled-over lists.
由于我不想编写代码来递增字母,因此我会创建一个包含所有有效 ID(AAAAAA 到 ZZZZZZ)的表,并为这些 ID 使用从 1 到 X 的整数。 然后您可以使用以下命令:
COALESCE 存在,以防表为空。 为了真正稳健,您应该计算“AAAAAA”(SELECT @min_silly_id =silly_id WHERE id = 1),以防您的“编号”算法发生变化。
如果您确实想把事情做好,您可以按照建议重新进行数据库设计。
Since I don't feel like writing code to increment letters, I'd create a table of all valid IDs (AAAAAA through ZZZZZZ) with an integer from 1 to X for those IDs. Then you can use the following:
The COALESCE is there in case the table is empty. To be truly robust you should calculate the 'AAAAAA' (SELECT @min_silly_id = silly_id WHERE id = 1) in case your "numbering" algorithm changes.
If you really wanted to do things right, you'd redo the database design as has been suggested.
我认为满足我的需求影响最小的解决方案是添加一个标识列。 我可以保证的一件事是,排序将首先添加应该“首先出现”的条目——我永远不会添加带有标识符 BBBB 的条目,然后再返回并添加 BBBA。 如果我没有这个约束,显然它不会工作,但就目前情况而言,我可以通过标识列进行排序并获得我想要的排序。
我会继续思考其他建议——也许如果它们在我脑海中“响起”,它们看起来会是一个更好的选择。
I think the lowest-impact solution for my needs is to add an identity column. The one thing I can guarantee is that the ordering will be such that entries that should "come first" will be added first -- I'll never add one with identifier BBBB, then go back and add BBBA later. If I didn't have that constraint, obviously it wouldn't work, but as it stands, I can just order by the identity column and get the sort I want.
I'll keep thinking about the other suggestions -- maybe if they "click" in my head, they'll look like a better option.
要返回给定
ID
的下一个ID
(带翻转),请使用:此查询搜索给定 ID 旁边的
ID
。 如果没有这样的ID
,则返回第一个ID
。结果如下:
,即对于
AAA
返回BBB
,对于BBB
返回CCC
等,并且,最后,AAA
代表EEE
,它位于表中的最后。To return the next
ID
for a givenID
(with rollover), use:This query searches for the
ID
next to the given. If there is no suchID
, it returns the firstID
.Here are the results:
, i. e. it returns
BBB
forAAA
,CCC
forBBB
, etc., and, finally,AAA
forEEE
which is last in the table.