Python 如何包装标签的所有内容?
我有以下XML部分:Python 如何包装标签的所有内容?,python,xml,xslt,Python,Xml,Xslt,我有以下XML部分: <table> <tr> <td>Hello</td> <td>Hello</td> <td> <p>Hello already in P</p> </td> <td> This one has some naked text <span>and som
<table>
<tr>
<td>Hello</td>
<td>Hello</td>
<td>
<p>Hello already in P</p>
</td>
<td>
This one has some naked text
<span>and some span wrapped text</span>
</td>
</tr>
</table>
我想在一个p标签中包装每个单元格的内容,这些内容还没有包装在一个p标签中。因此,输出为:
<table>
<tr>
<td><p>Hello</p></td>
<td><p>Hello</p></td>
<td>
<p>Hello already in p tag</p>
</td>
<td>
<p>
This one has some text
<span>and some span wrapped text</span>
</p>
</td>
</tr>
</table>
我在项目中使用lxml etree,但该库似乎没有wrap方法或类似方法
现在我想这可能是XSLT转换的工作,但我希望避免在Python项目中添加另一层复杂性和其他依赖项
td的内容可以是任何深度的我自己不使用lxml包,但请尝试以下方法:
def wrap(root):
# find <td> elements that do not have a <p> element
cells = etree.XPath("//td[not(p)]")(root)
for cell in cells:
# Create new <p> element
e = Element("p")
# Set the <p> element text from the parent
e.text = cell.text
# Clear the parent text because it is now in the <p> element
cell.text = None
# Move the parents children and make them the <p> element's children
# (because the span on line 10 of the input file should be nested)
for child in cell.getchildren():
# This actually moves the child from the <td> element to the <p> element
e.append(child)
# Set the new <p> element as the cell's child
cell.append(e)
嗨,谢谢你的回答。这个解决方案的问题是,它没有考虑到td的内容可以是任何深度的。你能给我一个新的示例输入文件和输出文件,上面的代码失败了,我会帮你修复它吗?事实上,我错了,你的答案对我有效。谢谢