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如何在Python中使用Unittest断言整数在范围内?_Python_Python Unittest - Fatal编程技术网

如何在Python中使用Unittest断言整数在范围内?

如何在Python中使用Unittest断言整数在范围内?,python,python-unittest,Python,Python Unittest,我有一个文件来验证标签 #configmanager.py class FunctionNode(object): def __validate_func_label(self, func_label): pattern = re.compile('^fn[1-9][0-9]{0,1}$') if not pattern.match(func_label): raise Exception( f"I

我有一个文件来验证标签

#configmanager.py

class FunctionNode(object):
    def __validate_func_label(self, func_label):
        pattern = re.compile('^fn[1-9][0-9]{0,1}$')
        if not pattern.match(func_label):
            raise Exception(
                f"Invalid value {func_label} for function label in function config.")

我必须使用单元测试为它编写一个测试用例。测试用例将检查用户输入的
fn
是否在1和99之间,它将通过。否则,如果
fn
超出范围,例如
fn0
fn100
则失败。我写了一个测试类

#test__validate_func_label.py

import unittest
from configmanager import FunctionNode

class TestValidateFuncLabel(unittest.TestCase):
    def setUp(self):
        pass

    def test_fn_within_range(self):
        for i in range(1,99):
            self.assertEqual(fn(i),output[i])

if __name__ == "__main__":
    unittest.main()
但我觉得我没有走上正轨。 是否有任何方法可以断言
fn
在1到99之间

感谢您的帮助。

self.assertTrue(0另一种解决方案是:

self.assertTrue( 0<=y<=99)

self.assertIn(fn, range(1, 100))