xml.etree.ElementTree:如何使用python访问具有属性名的节点

Question

需要帮助查找特定的"xsd:complexType name= 'Fort'">

我试着:

for a in root.findall('xsd:complexType name= 'Fort''):
print(a.attrib)

上面的代码不起作用

下面是xml:

<xsd:complexType name="Fort">
<xsd:sequence>
<xsd:element name="register" type="Skipsregister" />
<xsd:element name="type" type="pstype" />
<xsd:element name="om" type="aade" />
</xsd:sequence>
</xsd:complexType>

Answer 1

来自xml.etree.ElementTree文档:我们初始化一个树，然后通过tree. geroot()获取根节点。

那么你可以这样做:

for child in root:
if(child.attrib == "Fort"): #or child.tag, depending on what you mean by attribute
#do whatever you want here