Edu*_*scu 5 python xml minidom python-2.7
我有这种XML结构(从JSON转换而来的Esprima ASL的输出),它可以比this(ASL.xml)更加嵌套:
<?xml version="1.0" encoding="UTF-8" ?>
<program>
<type>Program</type>
<body>
<type>VariableDeclaration</type>
<declarations>
<type>VariableDeclarator</type>
<id>
<type>Identifier</type>
<name>answer</name>
</id>
<init>
<type>BinaryExpression</type>
<operator>*</operator>
<left>
<type>Literal</type>
<value>6</value>
</left>
<right>
<type>Literal</type>
<value>7</value>
</right>
</init>
</declarations>
<kind>var</kind>
</body>
</program>
for node inUsualy for XML我使用root.childNodes`但这仅适用于直接子节点:
import xml.dom.minidom as md
dom = md.parse("ASL.xml")
root = dom.documentElement
for node in root.childNodes:
if node.nodeType == node.ELEMENT_NODE:
print node.tagName,"has value:", node.nodeValue:, "and is child of:",node.parentNode.tagName
无论有多少嵌套元素,我如何遍历XML的所有元素？
这可能是使用递归函数最好的方法.像这样的东西应该这样做,但我没有测试它,所以考虑它伪代码.
import xml.dom.minidom as md
def print_node(root):
if root.childNodes:
for node in root.childNodes:
if node.nodeType == node.ELEMENT_NODE:
print node.tagName,"has value:", node.nodeValue, "and is child of:", node.parentNode.tagName
print_node(node)
dom = md.parse("ASL.xml")
root = dom.documentElement
print_node(root)