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xml - Groovy:将 XML 元素从一个文档复制到另一个文档

转载 作者:数据小太阳 更新时间:2023-10-29 02:22:18 26 4
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我是 Groovy 的新手,遇到了一个简单的问题。我想做的就是从一个 XML 文件中提取某些元素并用它创建一个新文件。这是一个示例 XML,让我们使用 Maven pom 文件:

<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>

我知道如何在 Groovy 中解析 XML:

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")
}

但是,我似乎无法将两者结合起来。例如,我想要获取 groupIdartifactId 和整个 dependencies 树,并从中创建一个新的 XML。没那么难,我想利用 Groovy 的简单性。

类似的东西(当然这是行不通的):

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

谢谢。该代码有很大帮助,但我如何处理命名空间,即如果输入中的项目标签如下所示:

<project xmlns="http://maven.apache.org/POM/4.0.0" 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">

然后它会在输出中添加一些奇怪的注释。我想要的只是输出中的项目标签看起来也像那样。

最佳答案

您可以使用 XmlSlurper 执行此操作:

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 )

打印:

<?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>

为了更好地处理命名空间,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>
| <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 )

哪个应该给你:

<?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>

关于xml - Groovy:将 XML 元素从一个文档复制到另一个文档,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/13715779/

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