如何将 XML 字符串转换为字典？

Question

我有一个从套接字读取 XML 文档的程序。我将 XML 文档存储在一个字符串中，我想将它直接转换为 Python 字典，就像在 Django 的 simplejson 库中完成的那样。

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 } }

原文由 user361526 发布，翻译遵循 CC BY-SA 4.0 许可协议

Answer 1

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

这是来自网站的代码，以防万一链接出错。

 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})

示例用法：

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

//或者，如果你想使用 XML 字符串：

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

原文由 James 发布，翻译遵循 CC BY-SA 4.0 许可协议

Answer 2

xmltodict （完全公开：我写的）正是这样做的：

 xmltodict.parse("""
<?xml version="1.0" ?>
<person>
  <name>john</name>
  <age>20</age>
</person>""")
# {u'person': {u'age': u'20', u'name': u'john'}}

原文由 Martin Blech 发布，翻译遵循 CC BY-SA 3.0 许可协议