blocks|key|852384|text|一种方法可能是使用python+sets模块，然后只做一个交集：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|852385|创建两个在id=5处重叠的查询集：|852386|In+[42]:+first+=+Location.objects.filter(id__lt=6)
In+[43]:+last+=+Location.objects.filter(id__gt=4)|code-block|syntax|javascript|852387|"import+sets“首先(得到一个弃用警告...嗯..。哦，好吧)。现在构建并相交它们-我们得到集合中的一个元素：|852388|In+[44]:+sets.Set(first).intersection(sets.Set(last))
Out[44]:+Set([<Location:+Location+object>])|852389|现在获取交叉点元素的id来检查它是否真的是5：|852390|In+[48]:+[s.id+for+s+in+sets.Set(first).intersection(sets.Set(last))]
Out[48]:+[5]|852391|这显然两次命中数据库并返回查询集的所有元素-更好的方法是将过滤器链接到管理器上，这应该能够在一次数据库命中中在SQL级别上完成。我看不到QuerySet.and/or(QuerySet)方法。|852392|entityMap^0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|V|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|W|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|X|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|Y|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|Z|8|@]|9|@]|A|$]]|$1|O|3|P|5|F|7|10|8|@]|9|@]|A|$G|H]]|$1|Q|3|R|5|6|7|11|8|@]|9|@]|A|$]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

One way may be to use the python sets module and just do an intersection:

make a couple of query sets that overlap at id=5:

<pre><code>In [42]: first = Location.objects.filter(id__lt=6)
In [43]: last = Location.objects.filter(id__gt=4)
</code></pre>

"import sets" first (gets a deprecation warning... ummm... oh well). Now build and intersect them - we get one element in the set:

<pre><code>In [44]: sets.Set(first).intersection(sets.Set(last))
Out[44]: Set([&lt;Location: Location object&gt;])
</code></pre>

Now get the id of the intersection elements to check it really is 5:

<pre><code>In [48]: [s.id for s in sets.Set(first).intersection(sets.Set(last))]
Out[48]: [5]
</code></pre>

This obviously hits the database twice and returns all the elements of the query set - better way would be to chain the filters on your managers and that should be able to do it in one DB hit and at the SQL level. I cant see a QuerySet.and/or(QuerySet) method.

blocks|key|852445|text|在大多数情况下，你可以直接写(利用QuerySet的"Set“部分)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|852446|intersection+=+Model.objects.filter(...)+&+Model.objects.filter(...)|code-block|syntax|javascript|852447|这并没有很好的文档记录，但是应该与在两个查询中的条件上使用AND条件几乎完全一样。相关代码：https://github.com/django/django/blob/1.8c1/django/db/models/query.py#L203|offset|length|852448|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/django/django/blob/1.8c1/django/db/models/query.py#L203^0|0|0|1A|22|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

In most cases you can just write (exploiting the "Set" part of QuerySet) :

<pre><code>intersection = Model.objects.filter(...) &amp; Model.objects.filter(...)
</code></pre>

This isn't very well documented, but should behave almost exactly like using AND conditions on conditions from both queries. Relevant code: <a href="https://github.com/django/django/blob/1.8c1/django/db/models/query.py#L203">https://github.com/django/django/blob/1.8c1/django/db/models/query.py#L203</a>

blocks|key|1066970|text|如果你想在python中完成，而不是在数据库中：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1066971|intersection+=+set(queryset1)+&+set(queryset2)|code-block|syntax|javascript|1066972|问题是，如果你在查询中使用不同的注释，由于添加了注释，对象看起来可能会不同……|1066973|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

If you want to do it in python, not in the database:

<pre><code>intersection = set(queryset1) &amp; set(queryset2)
</code></pre>

The problems is that if you use different annotations in the queriesdue to the added annotations the objects might look different...

blocks|key|852507|text|重构|type|unstyled|depth|inlineStyleRanges|entityRanges|data|852508|class+FeatureManager(models.Manager):

++++@staticmethod
++++def+_test_cases_eq_0(+qs+):
+++++++return+qs.annotate(+num_test_cases=models.Count('testcase_set')+).filter(num_test_cases=0)

++++@staticmethod
++++def+_standardized_gt_0(+qs+):
++++++++return+qs.annotate(+standardised=Count('documentation_set__standard')+).filter(standardised__gt=0)

++++def+without_test_cases(self):
++++++++return+self._test_cases_eq_0(+self.get_query_set()+)

++++def+standardised(self):
++++++++return+self._standardized_gt_0(+self.get_query_set()+)

++++def+intersection(+self+):
++++++++return+self._test_cases_eq_0(+self._standardized_gt_0(+self.get_query_set()+)+)|code-block|syntax|javascript|852509|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Refactor

<pre><code>class FeatureManager(models.Manager):

 @staticmethod
 def _test_cases_eq_0( qs ):
 return qs.annotate( num_test_cases=models.Count('testcase_set') ).filter(num_test_cases=0)

 @staticmethod
 def _standardized_gt_0( qs ):
 return qs.annotate( standardised=Count('documentation_set__standard') ).filter(standardised__gt=0)

 def without_test_cases(self):
 return self._test_cases_eq_0( self.get_query_set() )

 def standardised(self):
 return self._standardized_gt_0( self.get_query_set() )

 def intersection( self ):
 return self._test_cases_eq_0( self._standardized_gt_0( self.get_query_set() ) )
</code></pre>

blocks|key|1067076|text|如果你真的只是使用注解来根据计数是否为零进行过滤，那么这应该是可行的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1067077|class+FeatureManager(models.Manager):

++++def+without_test_cases(self):
++++++++return+self.get_query_set().filter(testcase__pk__isnull=True)

++++def+standardised(self):
++++++++return+self.get_query_set().filter(documentation_set__standard__isnull=False)|code-block|syntax|javascript|1067078|由于您不再担心注释，因此这两个查询应该会非常顺利地相交。|1067079|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

If you really are just using annotation to filter based on whether the count is zero or not, then this should work instead:

<pre><code>class FeatureManager(models.Manager):

 def without_test_cases(self):
 return self.get_query_set().filter(testcase__pk__isnull=True)

 def standardised(self):
 return self.get_query_set().filter(documentation_set__standard__isnull=False)
</code></pre>

Since you no longer are worrying about annotation, the two queries should intersect very smoothly.

blocks|key|1067120|text|我相信qs1.filter(pk__in=qs2)应该(通常)可以工作。对于我来说，它似乎也适用于类似的情况，它可以工作，并且生成的查询看起来很正常。(如果您的某个查询集使用value()来不选择主键列或其他奇怪的东西，我可以相信它会中断...)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1067121|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

I believe qs1.filter(pk__in=qs2) should work (usually). It seems to work for a similar case for me, it makes sense that it would work, and the generated query looks sane. (If one of your querysets uses values() to not select the primary key column or something weird, I can believe it'd break, though...)

blocks|key|852645|text|你可以这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|852646|intersection+=+queryset1+&+queryset2|code-block|syntax|javascript|852647|要进行联合，只需将&替换为%7C|offset|length|style|CODE|852648|entityMap^0|0|0|9|1|D|1|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]|$I|T|J|U|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|N|$]]

You can just do something like this:

<pre><code>intersection = queryset1 &amp; queryset2
</code></pre>

To do a union just replace <code>&amp;</code> by <code>|</code>

blocks|key|1067193|text|根据Django+1.11，现在可以使用函数intersection()|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1067194|>>>+qs1.intersection(qs2,+qs3)|code-block|syntax|javascript|1067195|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.djangoproject.com/en/1.11/ref/models/querysets/#intersection^0|M|E|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

As per Django 1.11, now it's available the function <a href="https://docs.djangoproject.com/en/1.11/ref/models/querysets/#intersection" rel="noreferrer">intersection()</a>

<pre><code>&gt;&gt;&gt; qs1.intersection(qs2, qs3)
</code></pre>

I’ve got a Django model with two custom manager methods. Each returns a different subset of the model’s objects, based on a different property of the object.

<pre><code>class FeatureManager(models.Manager):

 def without_test_cases(self):
 return self.get_query_set().annotate(num_test_cases=models.Count('testcase_set')).filter(num_test_cases=0)

 def standardised(self):
 return self.get_query_set().annotate(standardised=Count('documentation_set__standard')).filter(standardised__gt=0)
</code></pre>

(Both <code>testcase_set</code> and <code>documentation_set</code> refer to <code>ManyToManyField</code>s on other models.)

Is there any way to get a queryset, or just a list of objects, that’s the intersectiond of the querysets returned by each manager method?

How can I find the intersection of two Django querysets?

我有一个带有两个自定义管理器方法的Django模型。每种方法都根据对象的不同属性返回模型对象的不同子集。class FeatureManager(models.Manager):    def without_test_cases(self):        return self.get_query_set().an...

问如何找到两个Django查询集的交集？
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