我对Groovy还不熟悉,我只想解决一个简单的问题。我所要做的就是从一个XML文件中提取某些元素,并使用它创建一个新的文件。下面是一个XML示例,让我们使用一个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()
}我还知道如何在Groovy中创建XML:
def xml = new groovy.xml.MarkupBuilder()
xml.project (){
modelVersion("artifactId")
groupId("com.group")
artifactId("artifact")
}不过,我似乎很难将两者结合起来。例如,我希望以groupId、artifactId和整个依赖关系树为例,并从中创建一个新的XML。这不会很难,我想利用Groovy的简单性。
一些类似的东西(当然,这不起作用):
def newXml= new groovy.xml.MarkupBuilder()
newXml.groupId= project.groupId
newXml.dependencies = project.dependencyManagement.dependencies谢谢。该代码帮助很大,但是我如何处理名称空间,即如果输入中的项目标记看起来是这样的:
<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">然后将一些奇怪的注释添加到输出中。我想要的是输出中的项目标记也是这样。
发布于 2012-12-05 09:58:06
你可以用XmlSlurper来做这件事
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 )其中的指纹:
<?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>
| <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 )这应该会给你:
<?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>https://stackoverflow.com/questions/13715779
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