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Determine maximum number of columns from sqlite3

2019-04-30 19:55发布

问题:

Is it possible to get the maximum number of columns supported from sqlite3 at runtime? This database limitation is established with a compile-time variable SQLITE_MAX_COLUMN (see limits). The default is normally 2000 columns.

I'm looking for something accessible from either the Python or SQL interface.

回答1:

This appears to be impossible in practical terms (i.e. without a very expensive brute-force approach along the lines of dan04's rather brilliant answer).

The source (1, 2) for the sqlite3 module contains no reference either to SQLITE_MAX_COLUMN or to compile-time limits in general; neither does there appear to be any way to access them from within the SQL interface.

UPDATE:

A simple modification of dan04's solution to use a binary search speeds things up considerably:

import sqlite3

def max_columns():
    db = sqlite3.connect(':memory:')
    low = 1
    high = 32767  # hard limit <http://www.sqlite.org/limits.html>
    while low < high - 1:
        guess = (low + high) // 2
        try:
            db.execute('CREATE TABLE T%d (%s)' % (
                guess, ','.join('C%d' % i for i in range(guess))
            ))
        except sqlite3.DatabaseError as ex:
            if 'too many columns' in str(ex):
                high = guess
            else:
                raise
        else:
            low = guess
    return low

Running the code above through timeit.repeat():

>>> max_columns()
2000
>>> import timeit
>>> timeit.repeat(
...     "max_columns()", 
...     setup="from __main__ import max_columns",
...     number=50
... )
[10.347190856933594, 10.0917809009552, 10.320987939834595]

... which comes to an average run-time of 30.76 / 150 = 0.205 seconds (on a 2.6 GHz quad-core machine) - not exactly fast, but likely more usable than the 15-20 seconds of the "kick it 'til it breaks" counting-from-one method.



回答2:

A simple but inefficient way to do this from Python:

import itertools
import sqlite3

db = sqlite3.connect(':memory:')
try:
    for num_columns in itertools.count(1):
        db.execute('CREATE TABLE T%d (%s)' % (num_columns, ','.join('C%d' % i for i in range(num_columns))))
except sqlite3.DatabaseError as ex:
    if 'too many columns' in str(ex):
        print('Max columns = %d' % (num_columns - 1))