%{!?python_sitelib: %define python_sitelib %(python -c "from distutils.sysconfig import get_python_lib; print get_python_lib()")} Name: pycurl Version: 7.19.5.1 Release: 1%{?dist} Summary: A Python interface to libcurl Group: Development/Languages License: LGPLv2+ and an MIT/X URL: http://pycurl.sourceforge.net/ Source0: http://pycurl.sourceforge.net/download/pycurl-%{version}.tar.gz Vendor: VMware, Inc. Distribution: Photon Provides: pycurl Requires: python2 BuildRequires: openssl-devel BuildRequires: python2-devel BuildRequires: python2-libs BuildRequires: curl Requires: curl %description PycURL is a Python interface to libcurl. PycURL can be used to fetch objects identified by a URL from a Python program, similar to the urllib Python module. PycURL is mature, very fast, and supports a lot of features. %prep %setup -q -n pycurl-%{version} rm -f doc/*.xml_validity chmod a-x examples/* %build CFLAGS="$RPM_OPT_FLAGS -DHAVE_CURL_OPENSSL" python setup.py build %install rm -rf %{buildroot} python setup.py install -O1 --skip-build --root %{buildroot} rm -rf %{buildroot}%{_datadir}/doc/pycurl chmod 755 %{buildroot}%{python_sitelib}/pycurl.so %clean rm -rf %{buildroot} %files %defattr(-,root,root,-) %doc COPYING-LGPL COPYING-MIT RELEASE-NOTES.rst ChangeLog README.rst examples doc tests %{python_sitelib}/* %changelog * Sat Jan 24 2015 Touseef Liaqat <tliaqat@vmware.com> 7.19.5.1 - Initial build. First version