如何在使用XMLSlurper遍历文件时修改XML节点值 [英] How to modify XML node value, when iterating through the file with XMLSlurper

问题描述

$ b

  def root = new XmlSlurper（）。parseText（xml） 
 
 root.service.each {service  - > 
 service.receiver.endpoint.each {endpoint  - > 
 println\t\tEndpoint：$ {endpoint.text（）}
} 
} 
  

我想先读取endpoint.text（），然后修改节点的值，以便文件包含一个新值。
我正在阅读节点属性，因为它是。但是，我以后怎么写呢？

我看了这个问题，并了解如何写入现有文件。但是我问的是，当我已经遍历文件来读取它的一些内容时，它可以做得更加优雅吗？我希望有一个更有效的方式，适合我迭代文件的方式。

所以我希望有一种方法可以做到这一点：

p>

  def root = new XmlSlurper（）。parseText（xml）
 
 root.service.each {service  - > 
 service.receiver.endpoint.each {endpoint  - > 
 println\t\tEndpoint：$ {endpoint.text（）}
 // ****写入节点**** 
} 
} 
  

希望它有道理。

另外，这里有一些XML示例：

 < project命名= '普通' > 
< 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'> 
<发件人>发件人< /发件人> 
< attribute name ='operation'type ='String'> name< / attribute> 
< / operation> 
< / service> 
< / project> 
  

解决方案

不知道为什么 replaceBody 方法受到保护，但它可以工作：

  def root = new XmlSlurper（）。 parseText（
'''< root> 
< service> 
< receiver> 
< endpoint> 123< / endpoint> 
< endpoint> 456< / endpoint> 
< / receiver> 
< / service> 
< / root>'''）
 
 root.service.each {服务 - > 
 service.receiver.endpoint.each {endpoint  - > 
 endpoint.replaceBody（**+ endpoint.text（））
} 
} 
 
 println groovy.xml.XmlUtil.serialize（root）
  

I am currently iterating through an XML file like this:

def root = new XmlSlurper().parseText(xml)

root.service.each { service ->
  service.receiver.endpoint.each { endpoint ->
    println "\t\tEndpoint: ${endpoint.text()}"
    }
}

I want to first read the endpoint.text() , and afterwards modify the value of the node, so the file contains a new value. I am reading the node attributes fine, as it is. But how do I write to it afterwards?

I looked at this question, and understand how to write to an existing file. But what I am asking is, can it be done in a more elegant matter, when I am already iterating through the file to read some of it's content? I am hoping there is a more efficient way that suits the way I am iteraring through the file.

So am hoping there is a way to do something like this:

def root = new XmlSlurper().parseText(xml)

root.service.each { service ->
  service.receiver.endpoint.each { endpoint ->
    println "\t\tEndpoint: ${endpoint.text()}"
    // ****WRITE TO NODE****
    }
}

hope it makes sense.

Also, here is some example XML:

<project name='Common'>
  <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>
</project>

解决方案

don't know why replaceBody method is protected but it works:

def root = new XmlSlurper().parseText(
'''<root>
    <service>
        <receiver>
            <endpoint>123</endpoint>
            <endpoint>456</endpoint>
        </receiver>
    </service>
</root>''')

root.service.each { service ->
    service.receiver.endpoint.each { endpoint ->
        endpoint.replaceBody("**"+endpoint.text())
    }
}

println groovy.xml.XmlUtil.serialize( root )

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