Question

我正在用python编写一个类以从XML解析数据，我想将XML文件名作为参数传递，以便可以在构造函数中初始化树和根。我该怎么做？到目前为止，这是我的代码：

import xml.etree.ElementTree as ET

class cParser:

def __init__(self, file):
  tree = ET.parse(self.file)
  root = tree.getroot()
def getFilename():
   filename = root.attrib['filename']
   print("Filename is: %s" %(filename))

c1 = cParser('pythonxml.xml')
c1.getFilename()

Answer 1

尝试首先修正您的class声明：

import xml.etree.ElementTree as ET

class cParser:
    def __init__(self, file):
      tree = ET.parse(file)  # no need for self here
      self.root = tree.getroot()  # needs self here 

    def getFilename(self):  # missed self in arg list
       filename = self.root.attrib['filename']  # use self.root from init
       print("Filename is: %s" % filename)

c1 = cParser('pythonxml.xml')  # not a pythonic name for a class though
c1.getFilename()  # and not a pythonic name for method

如何在python方法中将XML文件作为参数传递？

1 个答案: