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1 ###############################################################################
2 # IPFire.org - An Open Source Firewall Solution #
3 # Copyright (C) - IPFire Development Team <info@ipfire.org> #
4 ###############################################################################
5
6 name = python-pycurl
7 version = 7.19.0
8 release = 3
9 thisapp = pycurl-%{version}
10
11 groups = Development/Languages
12 url = http://pycurl.sourceforge.net/
13 license = LGPLv2+ or MIT
14 summary = A Python interface to libcurl.
15
16 # During its initialization, PycURL checks that the actual libcurl version
17 # is not lower than the one used when PycURL was built.
18 # Yes, that should be handled by library versioning (which would then get
19 # automatically reflected).
20 # For now, we have to reflect that dependency.
21 LIBCURL_VERSION = $(shell curl-config --version | awk '{ print $NF }')
22 PKG_DEPS += libcurl>=%{LIBCURL_VERSION}
23
24 description
25 PycURL is a Python interface to libcurl. PycURL can be used to fetch
26 objects identified by a URL from a Python program, similar to the
27 urllib Python module. PycURL is mature, very fast, and supports a lot
28 of features.
29 end
30
31 source_dl =
32
33 build
34 requires
35 libcurl-devel
36 openssl-devel
37 python-devel
38 end
39
40 CFLAGS += -DHAVE_CURL_OPENSSL
41
42 build
43 python setup.py build
44 end
45
46 test
47 PYTHONPATH=$PWD/build/lib* python tests/test_internals.py -q
48 end
49
50 install
51 python setup.py install --skip-build -O1 \
52 --root=%{BUILDROOT}
53 rm -rf %{BUILDROOT}/usr/share/doc/pycurl
54 end
55 end
56
57 packages
58 package %{name}
59 end