Question

我要测试的派生代码中有此代码。

tigerContracts  = ['Element1','Element2']
def getReportingID(tigerKey):
    dig = '2' if tigerKey in tigerContracts else '0'
    return dig

我的单元测试是这样的：

class GeneratorDerived(unittest.TestCase):

    def tigerContracts(self):
        return ['Element3']

    @patch('derived.tigerContracts')
    def test_getReportingID(self, mockFunc):

        mockFunc.return_value = self.tigerContracts()

        reportingID = derived.getReportingID('XX-XX')
        self.assertEqual(reportingID,'0')

        reportingID = derived.getReportingID('Element3')
        self.assertEqual(reportingID,'2')

我的问题是，我正在将该方法作为返回值。
MagicMock name ='tigerContracts'id ='138171616'

相反，我希望返回值为
['Element3']

Answer 1

也许这是您正在寻找的更多信息，以得到解释：当我们调用self.tigerContracts（）时，将创建并返回一个新的模拟。根据我的评论：Full explanation

import unittest
import derived
from mock import patch
class GeneratorDerived(unittest.TestCase):

    def test_getReportingID(self):
        with patch('derived.tigerContracts', 'Element3'):

                reportingID = derived.getReportingID('XX-XX')
                self.assertEqual(reportingID,'0')

                reportingID = derived.getReportingID('Element3')
                self.assertEqual(reportingID,'2')

在py文件上模拟属性，以返回值而不是方法

1 个答案: