Question

我有一个sqlite表，其中3列名为ID（整数），N（整数）和V（实数）。该对（ID，N）是唯一的。

使用python模块sqlite3，我想用表单

执行递归选择

select ID from TABLE where N = 0 and V between ? and ? and ID in 
    (select ID from TABLE where N = 7 and V between ? and ? and ID in
        (select ID from TABLE where N = 8 and V between ? and ? and ID in  
            (...)
        )
    )

我收到以下错误，可能是因为超出了最大递归深度（？）。我需要大约20到50个回归水平

sqlite3.OperationalError: parser stack overflow

我也尝试加入像

这样的子选择

select ID from
         (select ID from TABLE where N = 0 and V between ? and ?) 
    join (select ID from TABLE where N = 7 and V between ? and ?) using (ID)
    join (select ID from TABLE where N = 8 and V between ? and ?) using (ID)
    join ...

但这种方法非常缓慢，即使只有很少的子选择

有没有更好的方法来执行相同的选择？
注意：该表在（N，V）上编入索引

以下是展示选择如何运作的示例

ID  N   V  
0   0   0,1  
0   1   0,2  
0   2   0,3  
1   0   0,5  
1   1   0,6  
1   2   0,7  
2   0   0,8  
2   1   0,9  
2   2   1,2

第0步

select ID from TABLE where N = 0 and V between 0 and 0,6

ID在（0,1）
中第1步

select ID from TABLE where N = 1 and V between 0 and 1  and ID in (0, 1)

ID仍在（0,1）
中第2步

select ID from TABLE where N = 2 and V between 0,5 and 1 and ID in (0, 1)

ID为1

Answer 1

打开递归，以相反的顺序执行，并在Python中执行。为此，我创建了一个由100条记录组成的表，每条记录的Id介于0到99之间，N = 3且V = 5。我随意选择了整个记录集作为最里面的记录。

您需要设想索引N和V的值列表，以便为最后一个SQL SELECT选择列表头部的值。循环所做的只是获取内部SELECT生成的ID列表，将其作为IN子句的一部分提供给下一个SELECT。

没有任何索引，这在augenblick中全部结束。

>>> import sqlite3
>>> conn = sqlite3.connect('recur.db')
>>> c = conn.cursor()
>>> previous_ids = str(tuple(range(0,100)))
>>> for it in range(50):
...     rows = c.execute('''SELECT ID FROM the_table WHERE N=3 AND V BETWEEN 2 AND 7 AND ID IN %s''' % previous_ids)
...     previous_ids = str(tuple([int(_[0]) for _ in rows.fetchall()]))
...     
>>> previous_ids
'(0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99)'

编辑：这可以避免使用长字符串，比augenblick需要更长的时间。它基本上是使用表格实现的相同想法。

>>> import sqlite3
>>> conn = sqlite3.connect('recur.db')
>>> c = conn.cursor()
>>> N_V = [
... (0, (0,6)),
... (0, (0, 1)),
... (1, (0, 2)),
... (2, (0, 3)),
... (0, (0, 5)),
... (1, (0, 6)),
... (2, (0, 7)),
... (0, (0, 8)),
... (1, (0, 9)),
... (2, (1, 2))
... ]
>>> r = c.execute('''CREATE TABLE essentials AS SELECT ID, N, V FROM the_table WHERE N=0 AND V BETWEEN 0 AND 6''')
>>> for n_v in N_V[1:]:
...     r = c.execute('''CREATE TABLE next AS SELECT * FROM essentials WHERE essentials.ID IN (SELECT ID FROM the_table WHERE N=%s AND V BETWEEN %s AND %s)''' % (n_v[0], n_v[1][0], n_v[1][1]))
...     r = c.execute('''DROP TABLE essentials''')
...     r = c.execute('''ALTER TABLE next RENAME TO essentials''')
...

Answer 2

索引三元组（ID，N，V）而不仅仅是（N，V）双联，使得连接方法足够快以便被考虑

create index I on TABLE(ID, N, V)

然后

select ID from
         (select ID from TABLE where N = 0 and V between ? and ?) 
    join (select ID from TABLE where N = 7 and V between ? and ?) using (ID)
    join (select ID from TABLE where N = 8 and V between ? and ?) using (ID)
    join ...

Answer 3

此查询仅需要（N，V）索引用于子查询。单独的ID索引可能有助于外部查询：

select ID from t
where ID in (select ID from TABLE where N = 0 and V between ? and ?)
  and ID in (select ID from TABLE where N = 7 and V between ? and ?)
  and ID in (select ID from TABLE where N = 8 and V between ? and ?)
  ...

python sqlite3中的递归（级联）选择

3 个答案: