Python xml - 删除空格以对齐 xml 文档

Question

我有一个 MyXml.xml，其结构为:

<?xml version='1.0' encoding='utf-8'?>
<tag1 atrib1='bla' atrib1='bla' atrib1='bla' atrib1='bla'>
    <tag2 atrib = 'something'>
        <tag3 atrib = 'something'>
           <tag4 atrib = '..'>
           </tag4>
        </tag3>
        <tag5 atrib = 'important'><div><h1>ContentFrom **OldXml.xml** </h1></div>
        ...
        </tag5>
    </tag2>
 </tag1>

有谁知道如何使其成为这种形式(删除所有空格):

<?xml version='1.0' encoding='utf-8'?>
<tag1 atrib1='bla' atrib1='bla' atrib1='bla' atrib1='bla'>
<tag2 atrib = 'something'>
<tag3 atrib = 'something'>
<tag4 atrib = '..'>
<tag5 atrib = 'important'><div><h1>ContentFrom **OldXml.xml** </h1></div>
...

我已经尝试过这个但行不通:

# Read in the file to a DOM data structure.
original_document = minidom.parse("MyXml.xml")

# Open a UTF-8 encoded file, because it's fairly standard for XML.
stripped_file = codecs.open("New_MyXml.xml", "w", encoding="utf8")

# Tell minidom to format the child text nodes without any extra whitespace.
original_document.writexml(stripped_file, indent="", addindent="", newl="")

stripped_file.close()

编辑:

文件是通过FOR循环创建的，其中创建了元素，最后写入是这样完成的:

    dom = xml.dom.minidom.parseString(ET.tostring(root))
    xml_string = dom.toprettyxml()
    part1, part2 = xml_string.split('?>')
    with open("MyXml.xml", 'w') as xfile:
          xfile.write(part1 + 'encoding=\"{}\"?>\n'.format(m_encoding) + part2)
          xfile.close()

编辑在一行中打印整个文档的最新代码:

    dom = xml.dom.minidom.parseString(ET.tostring(root))
    xml_string = dom.toxml()
    part1, part2 = xml_string.split('?>')
    xmlstring = f'{part1} encoding="{m_encoding}"?>\n {part2}'
    with open("MyXml.xml", 'w') as xfile:
        for line in xmlstring.split("\n"):
          xfile.write(line.strip() + "\n")

Answer 1

如果您只是想去除空格，则根本不需要(或不需要)XML 解析器:

from pathlib import Path

inf = Path("my-input.xml")
with inf.open() as f, inf.with_name(f"stripped-{inf.name}").open("w") as g:
    for line in f:
        g.write(line.strip() + "\n")

Pathlib 只是在这里扮演 os.path、open 等的角色:如果你不喜欢它，你可以在没有它的情况下重写(但是 pathlib 比修改路径的文本字符串要优越得多，我相信你不会想要......)

如果您确实需要加载解析器，请在编写时使用完全相同的技巧，但逐行迭代解析器对象。

---

演示:

from tempfile import TemporaryFile

data = """<?xml version='1.0' encoding='utf-8'?>
<tag1 atrib1='bla' atrib1='bla' atrib1='bla' atrib1='bla'>
    <tag2 atrib = 'something'>
        <tag3 atrib = 'something'>
           <tag4 atrib = '..'>
           </tag4>
        </tag3>
        <tag5 atrib = 'important'><div><h1>ContentFrom **OldXml.xml** </h1></div>
        ...
        </tag5>
    </tag2>
 </tag1>"""

with TemporaryFile(mode="w+") as f, TemporaryFile(mode="w+") as g:
    f.write(data)
    f.seek(0)
    print("Before:")
    for line in f:
        print(line, end="")
        g.write(line.strip() + "\n")

    print("\n\nAfter:")
    g.seek(0)
    for line in g:
        print(line, end="")

编辑:

在你的情况下，有一个更简单的解决方案:根本不使用 toprettyxml，使用 toxml。 (更新:显然渲染时根本没有换行符)。但即使没有它，我们也可以做同样的事情:

xml_string = dom.toprettyxml()
part1, part2 = xml_string.split('?>')
xmlstring = f'{part1} encoding="{m_encoding}"?>\n {part2}'
with open("MyXml.xml", 'w') as xfile:
    for line in xmlstring.split("\n"):
        xfile.write(line.strip() + "\n")

但是我怀疑 toprettyxml(indent="") 会做同样的事情:

xml_string = dom.toprettyxml(indent="")
...
with open("MyFile.xml", "w") as f:
    f.write(xml_string)