spack/var/spack/repos/builtin/packages/py-pycurl/package.py

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# Copyright 2013-2019 Lawrence Livermore National Security, LLC and other
# Spack Project Developers. See the top-level COPYRIGHT file for details.
#
# SPDX-License-Identifier: (Apache-2.0 OR MIT)
from spack import *
class PyPycurl(PythonPackage):
"""PycURL is a Python interface to libcurl. PycURL can be used to fetch
objects identified by a URL from a Python program."""
homepage = "http://pycurl.io/"
url = "https://pypi.io/packages/source/p/pycurl/pycurl-7.43.0.tar.gz"
version('7.43.0', sha256='aa975c19b79b6aa6c0518c0cc2ae33528900478f0b500531dbcdbf05beec584c')
depends_on('python@2.6:')
depends_on('curl@7.19.0:')