python 根据模板生成Jenkins配置文件

Posted 2021-05-09

#!/usr/bin/env python3

from os import path, makedirs
from string import Template

if __name__ == '__main__':

    def generate_config_xml(job, template, mapping, **kwds):
        print("\nGenerating Jenkins job config file in %s..." % job)
        makedirs(job, exist_ok=True)
        with open(path.join(job, 'config.xml'), 'w') as f:
            print(Template(template).substitute(mapping, **kwds), file=f)

    inc_modules = {'core', 'webapi', 'client'}
    exc_modules = {'client'}

    base_dir = path.dirname(path.realpath(__file__))
    conf_dir = path.join(base_dir, 'config')

    with open(path.join(base_dir, 'config-module.xml')) as f:
        conf_tpl_module = f.read()
    with open(path.join(base_dir, 'config-module-release.xml')) as f:
        conf_tpl_modrel = f.read()

    for m in sorted(inc_modules - exc_modules):
        mapping = {'module': m}

        generate_config_xml(path.join(conf_dir, 'app-%s' % m),
                            conf_tpl_module, mapping)
        generate_config_xml(path.join(conf_dir, 'app-%s-release' % m),
                            conf_tpl_modrel.replace('-Dmaven.test.skip=true', '-Dmaven.test.skip.exec=true') \
                                if m == 'process-impl' else conf_tpl_modrel,
                            mapping)