本文整理汇总了Python中spack.spec.Spec.namespace方法的典型用法代码示例。如果您正苦于以下问题：Python Spec.namespace方法的具体用法？Python Spec.namespace怎么用？Python Spec.namespace使用的例子？那么恭喜您, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类spack.spec.Spec的用法示例。
在下文中一共展示了Spec.namespace方法的1个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的Python代码示例。
示例1: get_repository
# 需要导入模块: from spack.spec import Spec [as 别名]
# 或者: from spack.spec.Spec import namespace [as 别名]
def get_repository(args, name):
"""Returns a Repo object that will allow us to determine the path where
the new package file should be created.
Args:
args (argparse.Namespace): The arguments given to ``spack create``
name (str): The name of the package to create
Returns:
Repo: A Repo object capable of determining the path to the package file
"""
spec = Spec(name)
# Figure out namespace for spec
if spec.namespace and args.namespace and spec.namespace != args.namespace:
tty.die("Namespaces '{0}' and '{1}' do not match.".format(
spec.namespace, args.namespace))
if not spec.namespace and args.namespace:
spec.namespace = args.namespace
# Figure out where the new package should live
repo_path = args.repo
if repo_path is not None:
repo = Repo(repo_path)
if spec.namespace and spec.namespace != repo.namespace:
tty.die("Can't create package with namespace {0} in repo with "
"namespace {0}".format(spec.namespace, repo.namespace))
else:
if spec.namespace:
repo = spack.repo.get_repo(spec.namespace, None)
if not repo:
tty.die("Unknown namespace: '{0}'".format(spec.namespace))
else:
repo = spack.repo.first_repo()
# Set the namespace on the spec if it's not there already
if not spec.namespace:
spec.namespace = repo.namespace
return repo