XML中如何判断节点是否有子节点_XML判断节点是否有子节点的方法与示例

const xmlStr = `
<book>
  <title>JavaScript入门</title>
  <author><name>张三</name></author>
</book>
`;
<p>const parser = new DOMParser();
const xmlDoc = parser.parseFromString(xmlStr, "text/xml");
const bookNode = xmlDoc.querySelector("book");</p><p>if (bookNode.children.length > 0) {
console.log("book节点有子节点");
} else {
console.log("book节点没有子节点");
}</p>

import xml.etree.ElementTree as ET
<p>xml_data = """<root>
<item>内容</item>
</root>"""</p><p>root = ET.fromstring(xml_data)</p>
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                <h1>方法一：转换为列表判断长度</h1><p>if len(list(root)) > 0:
print("root有子节点")</p><h1>方法二：使用for循环检测</h1><p>has_children = False
for child in root:
has_children = True
break</p><p>if has_children:
print("root包含子节点")</p>

// 判断book元素是否有子元素
const result = xmlDoc.evaluate('count(book/*)', xmlDoc, null, XPathResult.NUMBER_TYPE, null);
if (result.numberValue > 0) {
  console.log("book有子元素");
}

from lxml import etree
<p>root = etree.fromstring(xml_data)
if root.xpath("count(*)") > 0:
print("节点存在子节点")</p>