Question

我正在尝试迭代几个XML文件来比较它们中的数据。（大数据样本）。由于项目设置，我宁愿在groovy脚本中这样做。

XML布局是这样的:(假数据）

<service name='name' pattern='something' isReliable='maybe'>
  <receiver name='name' isUsingTwoWaySsl='maybe' isWsRmDisabled='maybe' targetedByTransformation='maybe'>
    <endpoint name='local_tst01'>URL</endpoint>
    <endpoint name='local_tst02'>URL</endpoint>
    <endpoint name='local_tst03'>URL</endpoint>
    <environment name='dev' default='local_dev' />
    <environment name='tst01' default='test' />
    <environment name='tst02' default='local_tst02' />
  </receiver>
  <operation name='name'>
    <sender>sender</sender>
    <attribute name='operation' type='String'>name</attribute>
  </operation>
</service>

如何迭代所有'端点'子元素？

请记住，我有一个非常大的数据样本，并且更喜欢有点标准化的解决方案。因为我必须经历许多不同类型的子元素。

Answer 1

你可以像这样进行深度优先搜索：

def xml = '''<service name='name' pattern='something' isReliable='maybe'>
  <receiver name='name' isUsingTwoWaySsl='maybe' isWsRmDisabled='maybe' targetedByTransformation='maybe'>
    <endpoint name='local_tst01'>URL</endpoint>
    <endpoint name='local_tst02'>URL</endpoint>
    <endpoint name='local_tst03'>URL</endpoint>
    <environment name='dev' default='local_dev' />
    <environment name='tst01' default='test' />
    <environment name='tst02' default='local_tst02' />
  </receiver>
  <operation name='name'>
    <sender>sender</sender>
    <attribute name='operation' type='String'>name</attribute>
  </operation>
</service>'''

new XmlSlurper().parseText(xml)
                .'**'
                .findAll { it.name() == 'endpoint' }
                .each { node ->
                    println "Found node with attributes ${node.attributes()} and body ${node.text()}"
                }

如何在groovy脚本中遍历xml文件

1 个答案: