从现有的动态web项目迁移到gradle

Question

我有一个使用eclipse创建的java动态web项目。我通常只是通过eclipse创建一个war文件，然后将其部署到tomcat。

现在我想用Gradle来构建我的项目并创建war文件。有没有eclipse插件可以做到这一点？如果没有，我如何在现有项目中使用gradle？我现有的项目结构是通过eclipse创建动态web项目时得到的，我不想更改它。

我尝试过阅读教程，并使用Gradle插件将其转换为Gradle项目。至少有人能指出一个博客、教程或一种开始的方式吗？

   MyProject
    |java resources
    |--src
    |-- -- packages
    |-- -- .properties files
    |--test
    |-- -- packages
    |build
    |-- classes
    |WebContent
    |-- MetaINF
    |-- WEB-INF
    |-- -- lib // all my libraries are here. there is no specific repo for now
    |-- -- web.xml

我的build.gradle：

apply plugin: 'java'
apply plugin: 'eclipse-wtp'
apply plugin: 'war'


sourceCompatibility = 1.7
targetCompatibility = 1.7


version = '2.0'

war {

baseName = 'Gradle'
version = '1.2'
}
repositories {
mavenCentral()
}

repositories {
flatDir {
   dirs 'lib'
   }
}

dependencies {
compile group: 'commons-collections', name: 'commons-collections', version: '3.2'
testCompile group: 'junit', name: 'junit', version: '4.+'

compile files('./lib/myjar.jar')    
compile 'commons-codec:commons-codec:1.5'

compile 'commons-lang:commons-lang:2.6'
compile 'org.hibernate:hibernate-entitymanager:4.3.8.Final'
compile 'org.apache.axis2:axis2-kernel:1.6.2'
compile 'org.apache.axis2:axis2-adb:1.6.2'
compile 'com.sun.jersey:jersey-server:1.19'
compile 'com.sun.jersey:jersey-client:1.19'
compile 'log4j:log4j:1.2.17'
compile 'org.apache.axis2:axis2-transport-local:1.6.2'
compile 'org.apache.axis2:axis2-transport-http:1.6.2'
providedCompile 'javax.servlet:servlet-api:2.3'



testCompile 'junit:junit:4.9'
testCompile 'org.jmock:jmock:2.6.0'
testCompile 'org.jmock:jmock-junit4:2.6.0'
testCompile 'org.jmock:jmock-legacy:2.6.0'



}

test {
systemProperties 'property': 'value'
}

网状物xml

<?xml version="1.0" encoding="UTF-8"?>
<web-app id="WebApp_ID" version="2.5" xmlns="http://java.sun.com/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd">
 <display-name>Gradle</display-name>
<servlet>
    <servlet-name>HTTP REST Service</servlet-name>
    <servlet-class>com.sun.jersey.spi.container.servlet.ServletContainer</servlet-class>
    //in the above line It is showing a warning: servlet-class references to non-existent class "com.sun.jersey.spi.container.servlet.ServletContainer"
    <init-param>
       <param-name>com.sun.jersey.config.property.packages</param-name>
       <param-value>org.gradle</param-value>
   </init-param>

    <load-on-startup>1</load-on-startup>


</servlet>
<servlet-mapping>
    <servlet-name>HTTP REST Service</servlet-name>
    <url-pattern>/*</url-pattern>
</servlet-mapping>

Answer 1

这里有一个最小的build.gradle文件，它应该让你去。使用命令gradle war构建。

apply plugin: 'war'

// See http://gradle.org/docs/current/userguide/war_plugin.html
//   Section 26.5. Convention properties
webAppDirName = 'WebContent'

// See http://gradle.org/docs/current/userguide/java_plugin.html
//   Section 23.4.1. Changing the project layout
sourceSets {
    main {
        // where does the Java source code live?
        java {
            srcDir 'Java Resources/src'
        }

        // where do classpath resources like *.properties files live?
        resources {
            srcDir 'Java Resources/src'
        }
    }
}

// See http://gradle.org/docs/current/userguide/dependency_management.html
//   Section 51.4.4. File dependencies
dependencies {
    // Where do the JARs live on the filesystem?
    compile fileTree(dir: "${webAppDirName}/WEB-INF/lib", include: '*.jar')
}