Question

我在python中使用dicttoxml将dict转换为XML。

我需要将dict转换为XML属性。

例如：

字典

[
      {
           "@name":"Ravi",
           "@age":21,
           "college":"Anna University"
       }
]

输出XML

<Student name="Ravi" age=21>
  <college>Anna University</college>
</Student>

码

dicttoxml(dict, custom_root='Student', attr_type=False, root=True)

实际输出

<Student>
  <key name="name">Ravi</key>
  <key name="age">21</key>
  <college>Anna University</college>
</Student>

Answer 1

我可能会建议declxml（完全披露：我写了）。使用declxml，您可以创建一个名为 processor 的对象，该对象以声明方式定义XML的结构。您可以使用处理器来解析和序列化XML数据。 declxml用于与字典，对象和命名元组进行序列化。它处理元素的属性和数组，并执行基本验证。

import declxml as xml


student = {
    'name':'Ravi',
    'age':21,
    'college':'Anna University'
}

student_processor = xml.dictionary('Student', [
    xml.string('.', attribute='name'),
    xml.integer('.', attribute='age'),
    xml.string('college')
])

xml.serialize_to_string(student_processor, student, indent='    ')

产生所需的输出

<?xml version="1.0" ?>
<Student age="21" name="Ravi">
    <college>Anna University</college>
</Student>

Answer 2

dicttoxml目前尚不支持此版本，但该问题已经开放很长时间了。 https://github.com/quandyfactory/dicttoxml/issues/27

虽然如果你的需求不是那么复杂，你可以试试这个简单的串行器。

https://gist.github.com/reimund/5435343/

在此处找到： - Serialize Python dictionary to XML

Answer 3

我也在使用xmltodict，我认为Hisham的回答中存在语法错误，但是我无法发表评论，所以这里是：

import xmltodict

xmldata = {
       'Student': {
           '@name':'Ravi',
           '@age':21,
           "college":"Anna University"
           }
   }

print(xmltodict.unparse(xmldata,pretty=True))

输出

<?xml version="1.0" encoding="utf-8"?>
<Student name="Ravi" age="21">
        <college>Anna University</college>
</Student>

Answer 4

我有类似的要求将XML转换为dict，反之亦然。我使用了一个名为xmltodict的库。此lib可让您从dict转换为具有属性的xml。

字典

xmldata = [
  {
       "@name":"Ravi",
       "@age":21,
       "college":"Anna University"
   }

]

代码

import xmltodict

print(xmltodict.unparse(xmldata), pretty=True)

Answer 5

这里有一些相对简单的代码，可以根据对象类型做出决定。如果一个对象是一个字典，它的键的值是历史上在 C 或 Java 等语言中被视为“原始”的键被写为属性，否则创建一个子元素。如果对象是列表，则为每个列表项创建一个 li 元素。属于原语的项目作为元素文本写出，而作为字典的项目作为属性写出。

#! python3-64

from lxml import etree

import json
import typing

def json_obj_to_xml(parent_element: typing.Optional[etree.Element], new_element_name: str, obj: typing.Union[bool, float, int, str, dict, list]):
    """
    Recursively walk an object and return its XML representation.

    Args:
        parent_element (typing.Optional[etree.Element]): The element that will be the parent of the element that this
            function will create and return.

        new_element_name (str): The name of the element that will be created.

        obj (typing.Union[bool, float, int, str, dict, list]): The object to return XML for.  It is expected that all
            objects passed to this function can be represented in JSON.

    Returns:
        result (etree.Element): An XML element.

    """

    if parent_element is not None:
        new_element = etree.SubElement(parent_element, new_element_name)

    else:
        new_element = etree.Element(new_element_name)

    if type(obj) == dict:
        for key, value in obj.items():
            if type(value) in (dict, list):
                json_obj_to_xml(new_element, key, value)

            else:
                # Convert values to a string, make sure boolean values are lowercase
                new_element.attrib[key] = str(value).lower() if type(value) == bool else str(value)

    elif type(obj) == list:
        for list_item in obj:
            # List items have to have a name.  Here we borrow "li" from HTML which stands for list item.
            json_obj_to_xml(new_element, 'li', list_item)

    else:
        # Convert everything to a string, make sure boolean values are lowercase
        new_element.text = str(obj).lower() if type(obj) == bool else str(obj)

    return new_element

# Read JSON file into a dictionary
json_file = r'C:\Users\ubiquibacon\Desktop\some_json_file.json'
json_file_hndl = open(json_file)
json_dict = json.load(json_file_hndl)
json_file_hndl.close()

# Recursively walk the dictionary to create the XML
root_xml_element = json_obj_to_xml(None, 'root', json_dict)

# Write the XML file
xml_file = f'{json_file}.xml'
with open(xml_file, 'wb') as xml_file_hndl:
    xml_file_hndl.write(etree.tostring(root_xml_element, pretty_print=True, xml_declaration=True))

使用属性将dict转换为XML

5 个答案: