迭代所有Xpath结果 [英] Iterate over all Xpath results

问题描述

 ＃！/ usr / bin / groovy 
 
 import javax .xml.xpath。* 
 import javax.xml.parsers.DocumentBuilderFactory 
 $ b $ def testxml ='''
< Employee> 
< ID> ..< / ID> 
<电子邮件> ..< /电子邮件> 
< custom_1> foo< / custom_1> 
< custom_2>栏< / custom_2> 
< custom_3> base< / custom_3> 
< / Employee> 
'''
 
 def processXml（String xml，String xpathQuery）{
 def xpath = XPathFactory.newInstance（）。newXPath（）
 def builder = DocumentBuilderFactory。 newInstance（）。newDocumentBuilder（）
 def inputStream = new ByteArrayInputStream（xml.bytes）
 def records = builder.parse（inputStream）.documentElement 
 xpath.evaluate（xpathQuery，records）
 
 
 println processXml（testxml，'// * [starts-with（name（），custom）]'）
  

，而不是返回所有节点（我在Xpath表达式中提供了 // ），我只获得第一个节点。如何修改我的代码以显示所有匹配的节点？

/docs.oracle.com/javase/7/docs/api/javax/xml/xpath/package-summary.htmlrel =nofollow> http://docs.oracle.com/javase/7/docs/api /javax/xml/xpath/package-summary.html 你传递 evaluate 你想要什么，默认为字符串。所以请求一个NODESET：

pre $ x $ c $ xpath.evaluate（xpathQuery，records，XPathConstants.NODESET）

并遍历结果的 NodeList ：

  def result = processXml（testxml，'// * [starts-with（name（），custom）]'）
 result.length.times {
 println result.item（it）.textContent 
} 
  

I have this code:

#!/usr/bin/groovy

import javax.xml.xpath.*
import javax.xml.parsers.DocumentBuilderFactory

def testxml = '''
                <Employee>
                  <ID>..</ID>
                  <E-mail>..</E-mail>
                  <custom_1>foo</custom_1>
                  <custom_2>bar</custom_2>
                  <custom_3>base</custom_3>
                </Employee>
  '''

def processXml( String xml, String xpathQuery ) {
  def xpath = XPathFactory.newInstance().newXPath()
  def builder     = DocumentBuilderFactory.newInstance().newDocumentBuilder()
  def inputStream = new ByteArrayInputStream( xml.bytes )
  def records     = builder.parse(inputStream).documentElement
  xpath.evaluate( xpathQuery, records )
}

println processXml( testxml, '//*[starts-with(name(), "custom")]' )

and instead of returning all the nodes (I provided // in Xpath expression), I get only the first node. How can I modify my code to display all the matching nodes ?

解决方案

according to the docs http://docs.oracle.com/javase/7/docs/api/javax/xml/xpath/package-summary.html you pass evaluate what you want, which defaults to string. So request a NODESET:

xpath.evaluate( xpathQuery, records, XPathConstants.NODESET )

and iterate over the resulting NodeList:

def result = processXml( testxml, '//*[starts-with(name(), "custom")]' )
result.length.times{
        println result.item(it).textContent
}

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