Python xml - 删除空格以获取对齐的 xml 文档

Question

我有一个 MyXml.xml 结构：

<?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>

有没有人知道如何使它成为这种形式（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>
...

我试过了，但是不行：

# 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：

文件是通过创建元素的FOR循环创建的，最后是这样写的：

    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

如果你只是想去掉空格，你根本不需要（或想要）一个 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 在这里只是扮演os.path、open 等角色：如果你碰巧不喜欢它，你可以在没有它的情况下重写（但pathlib 比为路径处理文本字符串要好得多）我确定你不想...）

如果您确实需要使用解析器进行加载，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：

在您的情况下，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)