如何将xml字符串转换为字典？

  • Z时代
  • 2024-01-10
  • 分类：问答

我有一个程序可以从套接字读取xml文档。我将xml文档存储在一个字符串中，我想将其直接转换为Python字典，就像在Django的simplejson库中一样。

举个例子：

str ="<?xml version="1.0" ?><person><name>john</name><age>20</age></person"
dic_xml = convert_to_dic(str)

然后dic_xml看起来像{'person' : { 'name' : 'john', 'age' : 20 } }

回答：

这是某人创建的一个很棒的模块。我已经使用过几次了。 http://code.activestate.com/recipes/410469-xml-as-

dictionary/

这是网站上的代码，以防万一链接损坏。

from xml.etree import cElementTree as ElementTree
class XmlListConfig(list):
    def __init__(self, aList):
        for element in aList:
            if element:
                # treat like dict
                if len(element) == 1 or element[0].tag != element[1].tag:
                    self.append(XmlDictConfig(element))
                # treat like list
                elif element[0].tag == element[1].tag:
                    self.append(XmlListConfig(element))
            elif element.text:
                text = element.text.strip()
                if text:
                    self.append(text)
class XmlDictConfig(dict):
    '''
    Example usage:
    >>> tree = ElementTree.parse('your_file.xml')
    >>> root = tree.getroot()
    >>> xmldict = XmlDictConfig(root)
    Or, if you want to use an XML string:
    >>> root = ElementTree.XML(xml_string)
    >>> xmldict = XmlDictConfig(root)
    And then use xmldict for what it is... a dict.
    '''
    def __init__(self, parent_element):
        if parent_element.items():
            self.update(dict(parent_element.items()))
        for element in parent_element:
            if element:
                # treat like dict - we assume that if the first two tags
                # in a series are different, then they are all different.
                if len(element) == 1 or element[0].tag != element[1].tag:
                    aDict = XmlDictConfig(element)
                # treat like list - we assume that if the first two tags
                # in a series are the same, then the rest are the same.
                else:
                    # here, we put the list in dictionary; the key is the
                    # tag name the list elements all share in common, and
                    # the value is the list itself 
                    aDict = {element[0].tag: XmlListConfig(element)}
                # if the tag has attributes, add those to the dict
                if element.items():
                    aDict.update(dict(element.items()))
                self.update({element.tag: aDict})
            # this assumes that if you've got an attribute in a tag,
            # you won't be having any text. This may or may not be a 
            # good idea -- time will tell. It works for the way we are
            # currently doing XML configuration files...
            elif element.items():
                self.update({element.tag: dict(element.items())})
            # finally, if there are no child tags and no attributes, extract
            # the text
            else:
                self.update({element.tag: element.text})

用法示例：

tree = ElementTree.parse('your_file.xml')
root = tree.getroot()
xmldict = XmlDictConfig(root)

//或者，如果要使用XML字符串：

root = ElementTree.XML(xml_string)
xmldict = XmlDictConfig(root)

以上是 如何将xml字符串转换为字典？ 的全部内容， 来源链接： utcz.com/qa/409311.html

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