Question

我正在检查字符串是否以“ 4”开头，它将自己分类为签证卡。运行pytest -v后，返回语法错误。在我的validator.py中，我的函数如下所示：

def get_issuer(number: str) -> str:
    if number.startswith('4'):
        return 'Visa'

我正在单独的文件中对此进行测试：

from card_validator.validator import get_issuer

def test_get_issuer_visa():
    assert get_issuer("4343 4212 1435 1231") == "Visa"

错误是

      def get_issuer(number: str) -> str:
E                            ^
E   SyntaxError: invalid syntax

我找不到任何错误。有什么帮助吗？

编辑： pytest -v的FULL输出为

========================================================== test session starts ==========================================================
platform linux2 -- Python 2.7.15rc1, pytest-3.3.2, py-1.5.2, pluggy-0.6.0 -- /usr/bin/python2
cachedir: .cache
rootdir: /home/bs-094/Dev/validator, inifile:
collected 0 items / 1 errors                                                                                                            

================================================================ ERRORS =================================================================
_______________________________________________ ERROR collecting tests/test_validators.py _______________________________________________
/usr/lib/python2.7/dist-packages/_pytest/python.py:403: in _importtestmodule
    mod = self.fspath.pyimport(ensuresyspath=importmode)
/usr/lib/python2.7/dist-packages/py/_path/local.py:668: in pyimport
    __import__(modname)
/usr/lib/python2.7/dist-packages/_pytest/assertion/rewrite.py:213: in load_module
    py.builtin.exec_(co, mod.__dict__)
tests/test_validators.py:1: in <module>
    from card_validator.validator import get_issuer
E     File "/home/bs-094/Dev/validator/card_validator/validator.py", line 1
E       def get_issuer(number: str) -> str:
E                            ^
E   SyntaxError: invalid syntax
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! Interrupted: 1 errors during collection !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
======================================================== 1 error in 0.12 seconds ========================================================

Answer 1

您的脚本包含属于python3的语法。因此，仅适用于python3。

要在python2.7（和python3中）中执行脚本，可以如下删除函数签名中的冒号：

def get_issuer(number):
    if number.startswith('4'):
        return 'Visa'

Answer 2

正如@Mark Tolonen在评论中提到的那样，键入是3.5版中的新增内容 https://docs.python.org/3/library/typing.html

您的代码在3.7.3中运行良好。升级到3.5+应该可以解决您的问题。

使用断言Python测试时语法无效

2 个答案: