How to retrieve odd-numbered nodes using xpath?

To retrieve odd-numbered nodes, you can use the position() function and the mod operator in XPath to determine if a node’s position is odd.

Here is an example XPath expression used to retrieve nodes at odd positions.

//*[position() mod 2 = 1]

This expression will match all nodes, but only return nodes that are at odd positions. You can modify * as needed to select specific types of nodes.

Here is an example XML document and sample code using this XPath expression:

XML document:

<root>
  <node1>Node 1</node1>
  <node2>Node 2</node2>
  <node3>Node 3</node3>
  <node4>Node 4</node4>
  <node5>Node 5</node5>
  <node6>Node 6</node6>
</root>

Sample code (using the lxml library in Python):

from lxml import etree

# 加载XML文档
xml = """
<root>
  <node1>Node 1</node1>
  <node2>Node 2</node2>
  <node3>Node 3</node3>
  <node4>Node 4</node4>
  <node5>Node 5</node5>
  <node6>Node 6</node6>
</root>
"""
root = etree.fromstring(xml)

# 使用XPath获取奇数节点
odd_nodes = root.xpath("//*[position() mod 2 = 1]")

# 打印结果
for node in odd_nodes:
    print(node.tag, node.text)

Output:

node1 Node 1
node3 Node 3
node5 Node 5

The above example demonstrates how to use XPath to retrieve nodes at odd positions. You can modify the XML document and XPath expression as needed.