Python(和编码新手)在这里。我正在尝试根据目录中的文件列表生成XML文件。文件名的前两个字母对应一个新的字母国家代码,我也试图提取它。
我的预期格式如下:
<ROOT>
<BASIC/>
<FULL>
<INFO>
<server>filname</server>
<country>country</country>
<region/>
</INFO>
</FULL>
</ROOT>
我似乎能够生成XML文件,但我无法使用pycountry将两位数的国家/地区代码转换为国家/地区。有人可以建议一个可能的解决方案?对其余代码的任何评论也会有所帮助。
# -*- coding: utf-8 -*-
import lxml.etree as xml
import pycountry
import glob
import gettext
gettext.bindtextdomain('iso3166', pycountry.LOCALES_DIR)
_c = lambda t: gettext.dgettext('iso3166', t)
def createXML(outfile):
root = xml.Element("ROOT")
basic = xml.Element("BASIC")
full = xml.Element("FULL")
root.append(basic)
root.append(full)
# add file information
for filename in glob.glob("*.*"):
info = xml.Element("INFO")
server = xml.SubElement(info, "server")
server.text = filename
short = filename[:2]
country = xml.SubElement(info, "country")
def get_country(code):
return _c(pycountry.countries.get(alpha2=code).name)
country.text = get_country(short)
region = xml.SubElement(info, "region")
full.append(info)
print xml.tostring(root, pretty_print=True)
#save new XML
# tree = xml.ElementTree(root)
# with open(filename, "w") as fh:
# tree.write(fh)
#--------------------------------------------------------
if __name__ == "__main__":
createXML("info.xml")