Django QuerySet 差异方法不起作用

Question

我试图找出两个 QuerySet 之间的区别，对我来说重要的是，这个区别的答案是一个 QuerySet。因此，自然的解决方案是使用 Django 查询集的差异方法。但是，当尝试这样做时，我收到以下错误：

NotSupportedError: difference is not supported on this database backend.

这是我想要做的：

In [4]: type(small_qs)
Out[4]: django.db.models.query.QuerySet

In [5]: type(bigger_qs)
Out[5]: django.db.models.query.QuerySet

In [6]: bigger_qs.difference(small_qs)

重要提示：两个查询集来自同一个模型。

其他有用的信息：

• 使用 docker（数据库和 django）
• 数据库后端是 MariaDB (MySQL)
• Django 版本为 2.0.6
• MariaDB 版本为 10.3.8

这里是完整的输出：

Out[6]: ---------------------------------------------------------------- 
NotSupportedError                         Traceback (most recent call 
last)
/usr/local/lib/python3.6/site-packages/IPython/core/formatters.py in 
__call__(self, obj)
700                 type_pprinters=self.type_printers,
701                 deferred_pprinters=self.deferred_printers)
702             printer.pretty(obj)
703             printer.flush()
704             return stream.getvalue()

/usr/local/lib/python3.6/site-packages/IPython/lib/pretty.py in 
pretty(self, obj)
398                         if cls is not object \
399                                 and 
callable(cls.__dict__.get('__repr__')):
400                             return _repr_pprint(obj, self, cycle)
401
402             return _default_pprint(obj, self, cycle)

/usr/local/lib/python3.6/site-packages/IPython/lib/pretty.py in 
_repr_pprint(obj, p, cycle)
693     """A pprint that just redirects to the normal repr function."""
694     # Find newlines and replace them with p.break_()
695     output = repr(obj)
696     for idx,output_line in enumerate(output.splitlines()):
697         if idx:

/usr/local/lib/python3.6/site-packages/django/db/models/query.py in         
__repr__(self)
246
247     def __repr__(self):
248         data = list(self[:REPR_OUTPUT_SIZE + 1])
249         if len(data) > REPR_OUTPUT_SIZE:
250             data[-1] = "...(remaining elements truncated)..."

/usr/local/lib/python3.6/site-packages/django/db/models/query.py in 
__iter__(self)
270                - Responsible for turning the rows into model 
objects.
271         """
272         self._fetch_all()
273         return iter(self._result_cache)
274

/usr/local/lib/python3.6/site-packages/django/db/models/query.py in 
_fetch_all(self)
1177     def _fetch_all(self):
1178         if self._result_cache is None:
1179             self._result_cache = list(self._iterable_class(self))
1180         if self._prefetch_related_lookups and not 
self._prefetch_done:
1181             self._prefetch_related_objects()

/usr/local/lib/python3.6/site-packages/django/db/models/query.py in 
__iter__(self)
51         # Execute the query. This will also fill compiler.select, 
klass_info,
52         # and annotations.
53         results = 
compiler.execute_sql(chunked_fetch=self.chunked_fetch, 
chunk_size=self.chunk_size)
54         select, klass_info, annotation_col_map = (compiler.select, 
compiler.klass_info,
55                                                   
compiler.annotation_col_map)

/usr/local/lib/python3.6/site-packages/django/db/models/sql/compiler.py 
in execute_sql(self, result_type, chunked_fetch, chunk_size)
1053             result_type = NO_RESULTS
1054         try:
1055             sql, params = self.as_sql()
1056             if not sql:
1057                 raise EmptyResultSet

/usr/local/lib/python3.6/site-packages/django/db/models/sql/compiler.py 
in as_sql(self, with_limits, with_col_aliases)
452             if combinator:
453                 if not getattr(features, 
'supports_select_{}'.format(combinator)):
454                     raise NotSupportedError('{} is not supported 
on this database backend.'.format(combinator))
455                 result, params = self.get_combinator_sql(combinator, 
self.query.combinator_all)
456             else:

NotSupportedError: difference is not supported on this database backend.

如果需要更多数据，请告诉我。

Answer 1

根据this ticket django 不正式支持MariaDB。大多数功能显然都在那里，但在确定版本或区分 MySQL 和 MariaDB 方面似乎有些困难。我猜你看到的错误与此有关。

也就是说，根据您要执行的操作，difference 实际上可能不是您问题的“自然解决方案”。如果（我怀疑）你的两个查询集实际上在同一个模型上，那么做你想做的最简单的方法是使用exclude。大致如下：

bigger_qs.exclude(id__in=smaller_qs)

这不会产生相同的 SQL（difference 使用 EXCEPT 而exclude 使用 WHERE NOT），它应该产生相同的结果集（可能以不同的顺序） .

Answer 2

此代码可能会有所帮助：

check = Qs1.objects.all()
prg=[]
[prg.append(x.ref) for x in check]
difference = (Qs2.objects.exclude(ref__in=prg).values())