比较 django TestCase 中的查询集 [英] comparing querysets in django TestCase

问题描述

我有个很简单的看法如下

I have a very simple view as follows

def simple_view(request):
    documents = request.user.document_set.all()
    return render(request, 'simple.html', {'documents': documents})

为了在我的测试用例中测试上述视图，我有以下方法会出错.

To test the above view in my test case i have the following method which errors out.

Class SomeTestCase(TestCase):
    # ...
    def test_simple_view(self):
        # ... some other checks
        docset = self.resonse.context['documents']
        self.assertTrue(self.user.document_set.all() == docset) # This line raises an error
    # ...

我得到的错误是 AssertionError: False is not true.我已经尝试打印两个查询集，并且两者都完全相同.当两个对象相同时，为什么会返回 False ?有什么想法吗?

The error i get is AssertionError: False is not true. I have tried printing both the querysets and both are absolutely identical. Why would it return False when both the objects are identical ? Any Ideas ?

目前为了克服这个问题，我正在使用一种令人讨厌的检查长度的方法，如下所示:

Currently to overcome this, I am using a nasty hack of checking lengths as follows:

ds1, ds2 = self.response.context['documents'], self.user.document_set.all()
self.assertTrue(len([x for x in ds1 if x in ds2]) == len(ds1) == len(ds2)) # Makes sure each entry in ds1 exists in ds2

推荐答案

如果查询集对象是不同查询的结果，即使它们的结果中具有相同的值(比较 ds1.query 和 ds2.query).

The queryset objects will not be identical if they are the result of different queries even if they have the same values in their result (compare ds1.query and ds2.query).

如果你先将查询集转换为列表，你应该可以进行正常的比较(当然假设它们具有相同的排序顺序):

If you convert the query set to a list first, you should be able to do a normal comparison (assuming they have the same sort order of course):

self.assertEqual(list(ds1), list(ds2))

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