Question

假设我有一个应用程序必须使用两个不同的后端：backend_1和backend_2。我有两个测试模块。

模块test_backend_1.py:

@py.test.fixture(scope="session")
def backend_1():
    return connect_to_backend_1()

def test_contract_method_1(backend_1):
    result = run_contract_method_1()
    assert result == backend_1_method_1_get_data()

def test_contract_method_2(backend_1):
    result = run_contract_method_2()
    assert result == backend_1_method_2_get_data()

和test_backend_2.py：

@py.test.fixture(scope="session")
def backend_2():
    return connect_to_backend_2()

def test_contract_method_1(backend_2):
    result = run_contract_method_1()
    assert result == backend_2_method_1_get_data()

def test_contract_method_2(backend_2):
    result = run_contract_method_2()
    assert result == backend_2_method_2_get_data()

我想合并这两个模块，而是提供具有相同接口的两个不同版本的fixture，并按顺序对每个模块运行我的测试。像这样：

@py.test.fixture(scope="session")
def backend():
    return [connect_to_backend_1(), connect_to_backend_2()]

def test_contract_method_1(backend):
    result = run_contract_method_1()
    assert result == backend_method_1_get_data()

def test_contract_method_2(backend):
    result = run_contract_method_2()
    assert result == backend_method_2_get_data()

注意：我必须有两个灯具的N个版本（每种类型的后端一个）：我的应用程序和本机后端连接器。我如何用py.test做到这一点？

Answer 1

我认为参数化夹具对你很有用：

import pytest

@pytest.fixture
def backends():
    """Mapping of possible backend ids to their constructor functions."""
    return {
        1: connect_to_backend_1,
        2: connect_to_backend_2
    }

@pytest.fixture(scope="session", params=[1, 2])
def backend(request, backends):
    """Parametrized backend instance."""
    return backends[request.param]()

def test_contract_method_1(backend):
    result = run_contract_method_1()
    assert result == backend.get_data()

def test_contract_method_2(backend):
    result = run_contract_method_2()
    assert result == backend.get_data()

现在还没有，但至少如果你完整阅读，你会得到更好的理解：

https://bitbucket.org/pytest-dev/pytest/src/da9d03b1f91d63ec97a989804bacfc03204b0e12/doc/en/fixture.txt?at=default

Answer 2

我认为，在您的情况下，您可以使用pytest.mark.parametrize功能：

def backend1():
    return connect_to_backend_1()

def backend2():
    return connect_to_backend_2()

def backend3():
    return connect_to_backend_3()

..

def backendN():
    return connect_to_backend_N()


@pytest.mark.parametrize('backend', [
    backend1, backend2, backend3, 
    ..., backendN
])
def test_something(backend):
    result = run_contract_method_2()
    assert result == backend.method_2_get_data()

使用不同版本的夹具运行一个测试

2 个答案: