python - 通过 python lxml tree.xpath 解析 xml

Question

我尝试解析一个大文件。示例如下。我尝试采取 <Name> ，但我不能仅在没有此字符串的情况下才有效

<LevelLayout xmlns="http://schemas.datacontract.org/2004/07/ArcherTech.Common.Domain" xmlns:i="http://www.w3.org/2001/XMLSchema-instance">

 

xml2 = '''<?xml version="1.0" encoding="UTF-8"?>
<PackageLevelLayout>
<LevelLayouts>
    <LevelLayout levelGuid="4a54f032-325e-4988-8621-2cb7b49d8432">
                <LevelLayout xmlns="http://schemas.datacontract.org/2004/07/ArcherTech.Common.Domain" xmlns:i="http://www.w3.org/2001/XMLSchema-instance">
                    <LevelLayoutSectionBase>
                        <LevelLayoutItemBase>
                            <Name>Tracking ID</Name>
                        </LevelLayoutItemBase>
                    </LevelLayoutSectionBase>
                </LevelLayout>
            </LevelLayout>
    </LevelLayouts>
</PackageLevelLayout>'''

from lxml import etree
tree = etree.XML(xml2)
nodes = tree.xpath('/PackageLevelLayout/LevelLayouts/LevelLayout[@levelGuid="4a54f032-325e-4988-8621-2cb7b49d8432"]/LevelLayout/LevelLayoutSectionBase/LevelLayoutItemBase/Name')
print nodes

Answer 1

您的嵌套 LevelLayout XML 文档使用命名空间。我会用:

tree.xpath('.//LevelLayout[@levelGuid="4a54f032-325e-4988-8621-2cb7b49d8432"]//*[local-name()="Name"]')

将 Name 元素与较短的 XPath 表达式相匹配(完全忽略命名空间)。

另一种方法是使用前缀到命名空间的映射并在标签上使用它们:

nsmap = {'acd': 'http://schemas.datacontract.org/2004/07/ArcherTech.Common.Domain'}

tree.xpath('/PackageLevelLayout/LevelLayouts/LevelLayout[@levelGuid="4a54f032-325e-4988-8621-2cb7b49d8432"]/acd:LevelLayout/acd:LevelLayoutSectionBase/acd:LevelLayoutItemBase/acd:Name',
    namespaces=nsmap)