如何使 lxml 的解析器保留根元素之外的空格？

发布于 2024-07-24 03:28:54 字数 506 浏览 7 评论 0原文

我正在使用 lxml 来操作一些现有的 XML 文档，并且我想引入尽可能少的差异噪音。不幸的是，默认情况下 lxml.etree.XMLParser 不会保留文档根元素之前或之后的空格：

>>> xml = '\n    <etaoin>shrdlu</etaoin>\n'
>>> lxml.etree.tostring(lxml.etree.fromstring(xml))
'<etaoin>shrdlu</etaoin>'
>>> lxml.etree.tostring(lxml.etree.fromstring(xml)) == xml
False

使用 lxml 可以吗？底层libxml2支持吗？

原文

I am using lxml to manipulate some existing XML documents, and I want to introduce as little diff noise as possible. Unfortunately by default lxml.etree.XMLParser doesn't preserve whitespace before or after the root element of a document:

>>> xml = '\n    <etaoin>shrdlu</etaoin>\n'
>>> lxml.etree.tostring(lxml.etree.fromstring(xml))
'<etaoin>shrdlu</etaoin>'
>>> lxml.etree.tostring(lxml.etree.fromstring(xml)) == xml
False

Is this possible using lxml? Is it supported by the underlying libxml2?

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∞梦里开花 2024-07-31 03:28:55

我不知道有哪个 XML 库可以为您做到这一点。但如果您确实需要这样做，那么使用正则表达式听起来是一个不错的主意。

>>> xml = '\n    <etaoin>shrdlu</etaoin>\n'
>>> head, tail = re.findall(r"^\s*|\s*$", xml)[:2]
>>> root = etree.fromstring(xml)
>>> out = head + etree.tostring(root) + tail
>>> out == xml
True

I don't know of any XML library that will do it for you. But using a regex sounds like a decent idea if you really need to do this.

>>> xml = '\n    <etaoin>shrdlu</etaoin>\n'
>>> head, tail = re.findall(r"^\s*|\s*$", xml)[:2]
>>> root = etree.fromstring(xml)
>>> out = head + etree.tostring(root) + tail
>>> out == xml
True

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