import codecs, os, re, sys
from xml.sax.saxutils import escape
fn = sys.argv[1]
ged = codecs.open(fn, encoding="cp437")
xml = codecs.open(fn+".xml", "w", "utf8")
xml.write("""<?xml version="1.0"?>\n""")
xml.write("<gedcom>")
sub = []
for s in ged:
s = s.strip()
m = re.match(r"(\d+) (@(\w+)@ )?(\w+)( (.*))?", s)
if m is None:
print "Error: unmatched line:", s
level = int(m.group(1))
id = m.group(3)
tag = m.group(4)
data = m.group(6)
while len(sub) > level:
xml.write("</%s>\n" % (sub[-1]))
sub.pop()
if level != len(sub):
print "Error: unexpected level:", s
sub += [tag]
if id is not None:
xml.write("<%s id=\"%s\">" % (tag, id))
else:
xml.write("<%s>" % (tag))
if data is not None:
m = re.match(r"@(\w+)@", data)
if m:
xml.write(m.group(1))
elif tag == "NAME":
m = re.match(r"(.*?)/(.*?)/$", data)
if m:
xml.write("<forename>%s</forename><surname>%s</surname>" % (escape(m.group(1).strip()), escape(m.group(2))))
else:
xml.write(escape(data))
elif tag == "DATE":
m = re.match(r"(((\d+)?\s+)?(\w+)?\s+)?(\d{3,})", data)
if m:
if m.group(3) is not None:
xml.write("<day>%s</day><month>%s</month><year>%s</year>" % (m.group(3), m.group(4), m.group(5)))
elif m.group(4) is not None:
xml.write("<month>%s</month><year>%s</year>" % (m.group(4), m.group(5)))
else:
xml.write("<year>%s</year>" % m.group(5))
else:
xml.write(escape(data))
else:
xml.write(escape(data))
while len(sub) > 0:
xml.write("</%s>" % sub[-1])
sub.pop()
xml.write("</gedcom>\n")
ged.close()
xml.close()
我从mwhite的答案中提取了代码,对其进行了一点扩展(好的,不仅仅是一点点),并发布在github:http://github.com/dijxtra/simplepyged。我采纳了关于其他补充内容的建议:-)
几年前,作为larger project的一部分,我用Python编写了一个简单的GEDCOM-to-XML转换器。我发现以XML格式处理GEDCOM数据要容易得多(特别是下一步涉及XSLT时)。在
我现在还没有在线代码,所以我已经将模块粘贴到这条消息中。这对我有效;没有保证。希望这有帮助。在
我知道这个线程很旧,但我在我的搜索和这个项目中找到了它https://github.com/madprime/python-gedcom/
源是安静干净和非常功能。在
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