Groovy：将XML文档从一个文档复制到另一个文档 [英] Groovy: copy XML elements from one doc to another

问题描述

我是Groovy的新手，并且陷入了一个简单的问题。我想要做的就是从一个XML文件中提取某些元素，并用它创建一个新文件。下面是一个XML示例，让我们使用Maven pom文件：

 < project> 
< modelVersion> 4.0.0< / modelVersion> 
< groupId> com.group< / groupId> 
< artifactId>工件< / artifactId> 
< version> 1.4< / version> 
< dependencyManagement> 
<依赖关系> 
< dependency> 
< groupId> junit< / groupId> 
< artifactId> junit< / artifactId> 
< version> 4.8.2< / version> 
< scope> test< / scope> 
< /依赖关系> 
< /依赖关系> 
< / dependencyManagement> 
  

我知道如何解析XML在Groovy中：
$ b $ pre $ def project = new XmlParser（）。parse（pom.xml）
project.groupId .each {
println it.text（）
}


我也知道如何在Groovy中创建XML：

  def xml = new groovy.xml.MarkupBuilder（）
 xml.project ）{
 modelVersion（artifactId）
 groupId（com.group）
 artifactId（artifact）
} 
  

然而，我认为将这两者结合起来存在问题。例如，我想要 groupId ， artifactId 和整个依赖关系树，并从中创建一个新的XML。
这不是那么难，我想要使用Groovy的简单性。

有些东西（当然这不行）： / p>

  def newXml = new groovy.xml.MarkupBuilder（）
 newXml.groupId = project.groupId 
 newXml .dependencies = project.dependencyManagement.dependencies 
  

---

谢谢。这段代码有很多帮助，但是我该如何处理名称空间，即如果输入中的项目标签如下所示：

 << ; project xmlns =http://maven.apache.org/POM/4.0.0xmlns：xsi =http://www.w3.org/2001/XMLSchema-instancexsi：schemaLocation =http：/ /maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd\"> 
  

然后在输出中添加一些奇怪的注释。我想要的就是输出中的项目标签也是这样。 你可以用 XmlSlurper ：

  import groovy.xml。* 
 
 def pxml ='''< project> 
 | < modelVersion> 4.0.0< / modelVersion> 
 | <&的groupId GT; com.group< /&的groupId GT; 
 | < artifactId的>伪影< / artifactId的> 
 | <版本> 1.4< /版本> 
 | < dependencyManagement> 
 | <依赖性> 
 | <依赖性> 
 | <&的groupId GT; junit的< /&的groupId GT; 
 | < artifactId的> junit的< / artifactId的> 
 | <版本> 4.8.2< /版本> 
 | <范围>试验< /范围> 
 | < /依赖性> 
 | < /依赖性> 
 | < / dependencyManagement> 
 |< / project>'''stripMargin（）
 
 def p = new XmlSlurper（）。parseText（pxml）
 
 String nxml = new StreamingMarkupBuilder （）.bind {
 project {
 dependecyManagement {
 dependencies {
 mkp.yield p.dependencyManagement.dependencies.children（）
} 
} 
 
 
 
 println XmlUtil.serialize（nxml）
  

 <？xml version =1.0encoding =UTF-8？> 
< project> 
< dependecyManagement> 
<依赖关系> 
< dependency> 
< groupId> junit< / groupId> 
< artifactId> junit< / artifactId> 
< version> 4.8.2< / version> 
< scope> test< / scope> 
< /依赖关系> 
< /依赖关系> 
< / dependecyManagement> 
< / project> 
  

要更好地处理名称空间，您可以尝试：

  def pxml ='''< project xmlns =http://maven.apache.org/POM/4.0.0
 | xmlns：xsi =http://www.w3.org/2001/XMLSchema-instance
 | xsi：schemaLocation =http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd\"> 
 | < modelVersion> 4.0.0< / modelVersion> 
 | <&的groupId GT; com.group< /&的groupId GT; 
 | < artifactId的>伪影< / artifactId的> 
 | <版本> 1.4< /版本> 
 | < dependencyManagement> 
 | <依赖性> 
 | <依赖性> 
 | <&的groupId GT; junit的< /&的groupId GT; 
 | < artifactId的> junit的< / artifactId的> 
 | <版本> 4.8.2< /版本> 
 | <范围>试验< /范围> 
 | < /依赖性> 
 | < /依赖性> 
 | < / dependencyManagement> 
 |< / project>'''stripMargin（）
 
 def p = new XmlSlurper（）。parseText（pxml）
 
 String nxml = new StreamingMarkupBuilder （）.bind {
 mkp.declareNamespace（''：http://maven.apache.org/POM/4.0.0，
'xsi'：http：//www.w3 （'xsi：schemaLocation'：p。@ schemaLocation）{
 dependecyManagement {
 dependencies {
 mkp.yield p.dependencyManagement .dependencies.children（）
} 
} 
} 
} 
 
 println XmlUtil.serialize（nxml）
 
 < / p> 
 > ;？xml version =1.0encoding =UTF-8？> 
< project xmlns =http://maven.apache.org/POM/4.0.0xmlns：xsi =http://www.w3.org/2001/XMLSchema-instancexsi：schemaLocation =http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd\"> 
< dependecyManagement> 
<依赖关系> 
< dependency> 
< groupId> junit< / groupId> 
< artifactId> junit< / artifactId> 
< version> 4.8.2< / version> 
< scope> test< / scope> 
< /依赖关系> 
< /依赖关系> 
< / dependecyManagement> 
< / project> 
  
 
I am new to Groovy and am stuck with a simple problem. All I wanna do is extract certain elements from one XML file and created a new file with it. Here's an example XML, let's use a Maven pom file:
<project>
  <modelVersion>4.0.0</modelVersion>
  <groupId>com.group</groupId>
  <artifactId>artifact</artifactId>
  <version>1.4</version>
  <dependencyManagement>
    <dependencies>
      <dependency>
        <groupId>junit</groupId>
        <artifactId>junit</artifactId>
        <version>4.8.2</version>
        <scope>test</scope>
      </dependency>
    </dependencies>
 </dependencyManagement>


I know how to parse XML in Groovy:
def project = new XmlParser().parse("pom.xml")
project.groupId.each{
  println it.text()
}
I also know how to create XML in Groovy:
def xml = new groovy.xml.MarkupBuilder()
xml.project (){
  modelVersion("artifactId")
  groupId("com.group")
  artifactId("artifact")
}
However, I seem to have a problem combining the two. I want, for example, take groupId, artifactId and the whole dependencies tree and create a new XML from it.
It can't be that hard and I wanna make use of the Groovy simplicity.

Something along those lines (of course this does not work):
def newXml= new groovy.xml.MarkupBuilder()
newXml.groupId= project.groupId
newXml.dependencies = project.dependencyManagement.dependencies




Thanks. That code helped a lot but how can I handle the namespace, i.e. if the project tag in the input looks like that:
<project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd">
It is then adding some weird annotations to the output. All I want is that the project tag in the output looks like that as well.
解决方案
You can do this with XmlSlurper:
import groovy.xml.*

def pxml = '''<project>
             |  <modelVersion>4.0.0</modelVersion>
             |  <groupId>com.group</groupId>
             |  <artifactId>artifact</artifactId>
             |  <version>1.4</version>
             |  <dependencyManagement>
             |    <dependencies>
             |      <dependency>
             |        <groupId>junit</groupId>
             |        <artifactId>junit</artifactId>
             |        <version>4.8.2</version>
             |        <scope>test</scope>
             |      </dependency>
             |    </dependencies>
             |  </dependencyManagement>
             |</project>'''.stripMargin()

def p = new XmlSlurper().parseText( pxml )

String nxml = new StreamingMarkupBuilder().bind {
  project {
    dependecyManagement {
      dependencies {
        mkp.yield p.dependencyManagement.dependencies.children()
      }
    }
  }
}

println XmlUtil.serialize( nxml )
Which prints:
<?xml version="1.0" encoding="UTF-8"?>
<project>
  <dependecyManagement>
    <dependencies>
      <dependency>
        <groupId>junit</groupId>
        <artifactId>junit</artifactId>
        <version>4.8.2</version>
        <scope>test</scope>
      </dependency>
    </dependencies>
  </dependecyManagement>
</project>
To handle the namespaces better, you could try:
def pxml = '''<project xmlns="http://maven.apache.org/POM/4.0.0"
             |         xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
             |         xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd">
             |  <modelVersion>4.0.0</modelVersion>
             |  <groupId>com.group</groupId>
             |  <artifactId>artifact</artifactId>
             |  <version>1.4</version>
             |  <dependencyManagement>
             |    <dependencies>
             |      <dependency>
             |        <groupId>junit</groupId>
             |        <artifactId>junit</artifactId>
             |        <version>4.8.2</version>
             |        <scope>test</scope>
             |      </dependency>
             |    </dependencies>
             |  </dependencyManagement>
             |</project>'''.stripMargin()

def p = new XmlSlurper().parseText( pxml )

String nxml = new StreamingMarkupBuilder().bind {
  mkp.declareNamespace(    '':"http://maven.apache.org/POM/4.0.0",
                        'xsi':"http://www.w3.org/2001/XMLSchema-instance" )
  project( 'xsi:schemaLocation':p.@schemaLocation ) {
    dependecyManagement {
      dependencies {
        mkp.yield p.dependencyManagement.dependencies.children()
      }
    }
  }
}

println XmlUtil.serialize( nxml )
Which should give you:
<?xml version="1.0" encoding="UTF-8"?>
<project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd">
  <dependecyManagement>
    <dependencies>
      <dependency>
        <groupId>junit</groupId>
        <artifactId>junit</artifactId>
        <version>4.8.2</version>
        <scope>test</scope>
      </dependency>
    </dependencies>
  </dependecyManagement>
</project>


                        
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