在pytest中，如何跳过或xfail某些灯具？

Question

我有一个严重固定的测试功能，它在某些灯具输入时失败（因为它应该）。我怎样才能表明这一点？这就是我现在正在做的，也许还有更好的办法。我很新py.test，所以我会很感激任何提示。

下一部分是所有输入夹具。仅供参考，example_datapackage_path在conf.test

@pytest.fixture(params=[None, 'pooled_col', 'phenotype_col']) 
def metadata_key(self, request): 
    return request.param 

@pytest.fixture(params=[None, 'feature_rename_col']) 
def expression_key(self, request): 
    return request.param 

@pytest.fixture(params=[None, 'feature_rename_col']) 
def splicing_key(self, request): 
    return request.param 

@pytest.fixture 
def datapackage(self, example_datapackage_path, metadata_key, 
       expression_key, splicing_key): 
    with open(example_datapackage_path) as f: 
     datapackage = json.load(f) 
    datatype_to_key = {'metadata': metadata_key, 
         'expression': expression_key, 
         'splicing': splicing_key} 
    for datatype, key in datatype_to_key.iteritems(): 
     if key is not None: 
      resource = name_to_resource(datapackage, datatype) 
      if key in resource: 
       resource.pop(key) 
    return datapackage 

@pytest.fixture 
def datapackage_dir(self, example_datapackage_path): 
    return os.path.dirname(example_datapackage_path) 

定义而这里的测试本身。

def test_from_datapackage(self, datapackage, datapackage_dir): 
    import flotilla 
    from flotilla.external import get_resource_from_name 

    study = flotilla.Study.from_datapackage(datapackage, datapackage_dir, 
              load_species_data=False) 

    metadata_resource = get_resource_from_name(datapackage, 'metadata') 
    expression_resource = get_resource_from_name(datapackage, 
               'expression') 
    splicing_resource = get_resource_from_name(datapackage, 'splicing') 

    phenotype_col = 'phenotype' if 'phenotype_col' \ 
     not in metadata_resource else metadata_resource['phenotype_col'] 
    pooled_col = None if 'pooled_col' not in metadata_resource else \ 
     metadata_resource['pooled_col'] 
    expression_feature_rename_col = 'gene_name' if \ 
     'feature_rename_col' not in expression_resource \ 
     else expression_resource['feature_rename_col'] 
    splicing_feature_rename_col = 'gene_name' if \ 
     'feature_rename_col' not in splicing_resource \ 
     else splicing_resource['feature_rename_col'] 

    assert study.metadata.phenotype_col == phenotype_col 
    assert study.metadata.pooled_col == pooled_col 
    assert study.expression.feature_rename_col \ 
      == expression_feature_rename_col 
    assert study.splicing.feature_rename_col == splicing_feature_rename_col 

我想这样做是metadata_key，说，当参数是pooled_col或phenotype_col，它就会失败。我查看了pytest: Skip and xfail: dealing with tests that can not succeed，但它只讨论了skip和xfail进行参数化测试，但没有讨论固定装置。

Answer 1

在你datapackage或expression_key灯具描述here可以使用pytest.xfail和pytest.skip。例如：

@pytest.fixture 
def datapackage(self, example_datapackage_path, metadata_key, 
       expression_key, splicing_key): 
    if metadata_key == 'pooled_col': 
     pytest.skip('metadata key is "pooled_col"') 
    ... 

您还可以使用pytest.mark.xfail在夹具参数，在这个例子中：

@pytest.fixture(params=['a', pytest.mark.xfail('b'), 'c']) 
def fx1(request): 
    return request.param 


def test_spam(fx1): 
    assert fx1 

如果你喜欢跳过这些测试，这似乎工作：

@pytest.fixture(
    params=['a', pytest.mark.skipif(True, reason='reason')('b'), 'c']) 
def fx1(request): 
    return request.param 


def test_spam(fx1): 
    assert fx1