sqlite3+ManyToManyField: unable to link to 1000 (or more) objects.

[Jakub Wilk <django@…>]

With DATABASE_ENGINE = 'sqlite3' and the following foo/models.py:

from django.db import models

class Foo(models.Model):
        pass

class Bar(models.Model):
        foos = models.ManyToManyField(Foo)

I try:

>>> from foo import models
>>> models.Foo.objects.all().count()
1733
>>> bar = models.Bar(1)
>>> bar.foos = models.Foo.objects.all()[:1000]
Traceback (most recent call last):
  File "<console>", line 1, in ?
  File "/var/lib/python-support/python2.4/django/db/models/fields/related.py", line 460, in __set__
    manager.add(*value)
  File "/var/lib/python-support/python2.4/django/db/models/fields/related.py", line 287, in add
    self._add_items(self.source_col_name, self.target_col_name, *objs)
  File "/var/lib/python-support/python2.4/django/db/models/fields/related.py", line 338, in _add_items
    [self._pk_val] + list(new_ids))
  File "/var/lib/python-support/python2.4/django/db/backends/util.py", line 19, in execute
    return self.cursor.execute(sql, params)
  File "/var/lib/python-support/python2.4/django/db/backends/sqlite3/base.py", line 129, in execute
    return Database.Cursor.execute(self, query, params)
OperationalError: too many SQL variables

[James Bennett]

This feels like a built-in limit of SQLite; you may need to break this up into multiple operations.

[deepak <deep.thukral@…>]

Worked without any problem

>>> from abc.models import *
>>> for i in range(2000): Foo().save()
>>> print Foo.objects.all().count()
2000
>>> bar = Bar(1)
>>> bar.foos = Foo.objects.all()[:1000]
>>> bar.save()
>>>

I'd suggest you to post SQL queries fired by django.