Information for RPM python-pycurl-7.19.0-19.el7.src.rpm
ID | 3943 | |||||
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Build | python-pycurl-7.19.0-19.el7 | |||||
Name | python-pycurl | |||||
Version | 7.19.0 | |||||
Release | 19.el7 | |||||
Epoch | ||||||
Arch | src | |||||
Draft | False | |||||
Summary | A Python interface to libcurl | |||||
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. | |||||
Build Time | 2016-11-05 15:10:22 GMT | |||||
Size | 92.17 KB | |||||
cdea2358016e6b42ac01e251d3fddcea | ||||||
License | LGPLv2+ or MIT | |||||
Provides | No Provides | |||||
Obsoletes | No Obsoletes | |||||
Conflicts | No Conflicts | |||||
Requires |
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Recommends | No Recommends | |||||
Suggests | No Suggests | |||||
Supplements | No Supplements | |||||
Enhances | No Enhances | |||||
Files | ||||||
Component of | No Buildroots |