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File python-pycurl.spec of Package python-pycurl

# spec file for package python-xdg (Version 0.17)
# Copyright (c) 2009 SUSE LINUX Products GmbH, Nuernberg, Germany.
# All modifications and additions to the file contributed by third parties
# remain the property of their copyright owners, unless otherwise agreed
# upon. The license for this file, and modifications and additions to the
# file, is the same license as for the pristine package itself (unless the
# license for the pristine package is not an Open Source License, in which
# case the license is the MIT License). An "Open Source License" is a
# license that conforms to the Open Source Definition (Version 1.9)
# published by the Open Source Initiative.

# Please submit bugfixes or comments via http://bugs.opensuse.org/

Name:           python-pycurl
%define         _name pycurl
Summary:        Python library to access URLs
Version:        7.19.0
Release:        0
License:        LGPL v2.1 or later
Group:          Development/Libraries/Python
Url:            http://pycurl.sourceforge.net/
Source:         %{_name}-%{version}.tar.bz2
Patch:		suse.patch
BuildRequires:  fdupes
BuildRequires:  python-devel libcurl-devel
BuildRoot:      %{_tmppath}/%{name}-%{version}-build



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.

%setup -q -n %{_name}-%{version}
%patch -p1

%{__python} setup.py build

%{__python} setup.py install -O1 --skip-build --root="%{buildroot}" --prefix="%{_usr}" --record=INSTALLED_FILES

%fdupes %{buildroot}/%{python_sitelib}

%{__rm} -rf %{buildroot}

%defattr(-, root, root, 0755)
%dir %py_sitedir/curl
%dir /usr/share/doc/packages/python-pycurl
%dir /usr/share/doc/packages/python-pycurl/examples
%dir /usr/share/doc/packages/python-pycurl/html
%dir /usr/share/doc/packages/python-pycurl/tests

* Tue Mar 16 2010 admin@eregion.de
- initial package