使用LXML编写XML标头
发布时间:2020-12-16 23:14:18 所属栏目:百科 来源:网络整理
导读:我目前正在编写一个脚本,将一堆 XML文件从各种编码转换为统一的UTF-8. 我首先尝试使用LXML确定编码: def get_source_encoding(self): tree = etree.parse(self.inputfile) encoding = tree.docinfo.encoding self.inputfile.seek(0) return (encoding or ''
我目前正在编写一个脚本,将一堆
XML文件从各种编码转换为统一的UTF-8.
我首先尝试使用LXML确定编码: def get_source_encoding(self): tree = etree.parse(self.inputfile) encoding = tree.docinfo.encoding self.inputfile.seek(0) return (encoding or '').lower() 如果那是空白的,我尝试从chardet获取它: def guess_source_encoding(self): chunk = self.inputfile.read(1024 * 10) self.inputfile.seek(0) return chardet.detect(chunk).lower() 然后我使用编解码器转换文件的编码: def convert_encoding(self,source_encoding,input_filename,output_filename): chunk_size = 16 * 1024 with codecs.open(input_filename,"rb",source_encoding) as source: with codecs.open(output_filename,"wb","utf-8") as destination: while True: chunk = source.read(chunk_size) if not chunk: break; destination.write(chunk) 最后,我正在尝试重写XML标头.如果最初是XML标头 <?xml version="1.0"?> 要么 <?xml version="1.0" encoding="windows-1255"?> 我想把它变成 <?xml version="1.0" encoding="UTF-8"?> 我目前的代码似乎不起作用: def edit_header(self,input_filename): output_filename = tempfile.mktemp(suffix=".xml") with open(input_filename,"rb") as source: parser = etree.XMLParser(encoding="UTF-8") tree = etree.parse(source,parser) with open(output_filename,"wb") as destination: tree.write(destination,encoding="UTF-8") 我正在测试的文件有一个没有指定编码的标头.如何使用指定的编码正确输出标题? 解决方法
尝试:
tree.write(destination,xml_declaration=True,encoding='UTF-8') 从the API docs开始:
来自ipython的示例: In [15]: etree.ElementTree(etree.XML('<hi/>')).write(sys.stdout,encoding='UTF-8') <?xml version='1.0' encoding='UTF-8'?> <hi/> 经过反思,我觉得你太努力了. lxml会自动检测编码并根据该编码正确解析文件. 所以你真正要做的事情(至少在Python2.7中)是: def convert_encoding(self,output_filename): tree = etree.parse(input_filename) with open(output_filename,'w') as destination: tree.write(destination,encoding='utf-8',xml_declaration=True) (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |