在Django TestCase中比较查询集 [英] comparing querysets in django TestCase

问题描述

我有一个非常简单的视图，如下所示

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})

要在我的测试用例中测试以上视图，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：假不是真的。
我尝试打印两个查询集，并且两个都完全相同。当两个对象相同时，为什么返回 False ？有任何想法吗？

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 ?

当前为了克服这个问题，我正在使用一种讨厌的方法来检查长度，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

queryset对象是不同查询的结果，即使它们的结果相同（即使 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).

如果转换查询集首先到列表，您应该能够进行常规比较（假设它们当然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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