Python 解析 Spring pom 文件获取 jar 包名称

前段时间在做持续集成有个小需求是根据 pom 文件获取 jar 包名称，在网上搜寻一番，整理了一份脚本，可以直接使用，通过解析 pom 文件获取(xml2pydict.py)：
使用示例：python xml2pydict.py pom.xml
输出结果：jar 包名称

#!/usr/bin/env python
# -*- coding: utf-8 -*-
# @Time    : 2018/10/22 下午5:53
# @Author  : haohao.qiang
# @Mail    : qhh0205@gmail.com
# @File    : xml2pydict.py

import sys
import warnings
from xml.parsers import expat
import xml.etree.ElementTree as ElementTree
warnings.filterwarnings("ignore")

class XmlListConfig(list):
    def __init__(self, aList):
        for element in aList:
            if element:
                # treat like dict
                if len(element) == 1 or element[0].tag != element[1].tag:
                    self.append(XmlDictConfig(element))
                # treat like list
                elif element[0].tag == element[1].tag:
                    self.append(XmlListConfig(element))
            elif element.text:
                text = element.text.strip()
                if text:
                    self.append(text)


class XmlDictConfig(dict):
    '''
    Example usage:

    tree = ElementTree.parse('your_file.xml')
    root = tree.getroot()
    xmldict = XmlDictConfig(root)

    Or, if you want to use an XML string:

    root = ElementTree.XML(xml_string)
    xmldict = XmlDictConfig(root)

    And then use xmldict for what it is... a dict.
    '''
    def __init__(self, parent_element):
        if parent_element.items():
            self.update(dict(parent_element.items()))
        for element in parent_element:
            if element:
                # treat like dict - we assume that if the first two tags
                # in a series are different, then they are all different.
                if len(element) == 1 or element[0].tag != element[1].tag:
                    aDict = XmlDictConfig(element)
                # treat like list - we assume that if the first two tags
                # in a series are the same, then the rest are the same.
                else:
                    # here, we put the list in dictionary; the key is the
                    # tag name the list elements all share in common, and
                    # the value is the list itself
                    aDict = {element[0].tag: XmlListConfig(element)}
                # if the tag has attributes, add those to the dict
                if element.items():
                    aDict.update(dict(element.items()))
                self.update({element.tag: aDict})
            # this assumes that if you've got an attribute in a tag,
            # you won't be having any text. This may or may not be a
            # good idea -- time will tell. It works for the way we are
            # currently doing XML configuration files...
            elif element.items():
                self.update({element.tag: dict(element.items())})
            # finally, if there are no child tags and no attributes, extract
            # the text
            else:
                self.update({element.tag: element.text})


if __name__ == "__main__":
    xml = sys.argv[1]
    oldcreate = expat.ParserCreate
    expat.ParserCreate = lambda encoding, sep: oldcreate(encoding, None)
    tree = ElementTree.parse(xml)
    root = tree.getroot()
    xmldict = XmlDictConfig(root)
    print "{}-{}.{}".format(xmldict.get('artifactId'), xmldict.get('version'), xmldict.get('packaging'))