Question

我正在尝试在文件夹中的所有xml文件上运行我的代码我运行代码时会出现一些错误并生成一些文件但不是全部

这是我的代码：

import xml.etree.ElementTree as ET
import os
import glob
path = 'C:/xml/'

for infile in glob.glob( os.path.join(path, '*.xml') ):
        tree = ET.parse(infile)
        root = tree.getroot()
        with open(infile+'new.csv','w') as outfile:
            for elem in root.findall('.//event[@type="MEDIA"]'):
                    mediaidelem = elem.find('./mediaid')
                    if mediaidelem is not None:
                            outfile.write("{}\n".format(mediaidelem.text))

这是所有

的错误日志

Traceback (most recent call last):
  File "C:\xml\2.py", line 8, in <module>
    tree = ET.parse(infile)
  File "C:\Python34\lib\xml\etree\ElementTree.py", line 1187, in parse
    tree.parse(source, parser)
  File "C:\Python34\lib\xml\etree\ElementTree.py", line 598, in parse
    self._root = parser._parse_whole(source)
  File "<string>", line None
xml.etree.ElementTree.ParseError: no element found: line 1, column 0

Answer 1

考虑到错误消息，您可能会有一些空（或格式错误）的文件。

我会在此处添加错误处理以警告用户此类错误，然后跳过该文件。类似的东西：

for infile in glob.glob( os.path.join(path, '*.xml') ):
    try:
        tree = ET.parse(infile)
    except xml.etree.ElementTree.ParseError as e:
        print infile, str(e)
        continue
    ...

我没有尝试在这里重现它，这只是猜测。

python解析/处理文件夹中的所有xml文件

1 个答案: