Gradle：从 .class 文件创建 .jar 并在构建 EAR 时将其包含在 EAR_dir/lib 中

Question

你会认为这很容易。 Gradle/Maven 是专门为摆脱构建噩梦而设计的。然而......我已经搜索了网络，包括SO。我更喜欢使用 Maven，但这不是我能控制的。

我的主 bu​​ild.gradle 文件如下所示：

buildscript {
 repositories {
    maven { url "https://aaa.com/xxxx/aaa-mvn" }
}
dependencies {
    {redacted}
    classpath "com.aaa.plugin.gradle:module-plugin:1.+"
    classpath "org.jfrog.buildinfo:build-info-extractor-gradle:4.4.4"
 }
}

apply plugin: 'maven-publish'
apply plugin: {redacted}
apply plugin: {redacted}
apply plugin: 'build-defaults'
apply plugin: 'module-plugin'
apply plugin: 'war'

description = 'xxxxxx'

defaultTasks 'build','install'

dependencies {
 jbossModule(group: 'com.aaa.bbbb', name: 'inf-jdbc', version: '3.0.2')
}

artifactory {
 publish {
    repoKey=version.endsWith("SNAPSHOT") ? 'aaa-mvn-dist-snapshots' : 'aaa-mvn-dist'
    defaults {
        publications('mavenJava')
    }
 }
}

publishing {
 publications {
    mavenJava(MavenPublication) {
        artifact project(':xxxEAR').ear
    }
 }
}

repositories {
 maven { url "https://xxxx.aaa.com/artifactory/aaa-mvn" }
}

这是我的 .war 的 gradle.build。我在 .ear 中有一个 .war。

description = 'Pricing'

buildscript {
 repositories {
    maven { url "https://xxx.aaa.com/artifactory/aaa-mvn" }
 }
 dependencies {
    classpath "com.aaa.plugin.gradle:ucd-publish-plugin:1.+"
    classpath "com.aaa.plugin.gradle:build-defaults-plugin:1.+"
    classpath {redacted}
    classpath {redacted}
 }
}

apply plugin: 'java'
apply plugin: 'war'
apply plugin: 'jacoco'
apply plugin: {redacted}
apply plugin: 'ucd-publish'
apply plugin: 'build-defaults'
apply plugin: {redacted}
apply plugin: 'maven'
apply plugin: 'maven-publish'

repositories {
  maven { url "https://xxxx.aaa.com/artifactory/aaa-mvn" }
}

war {
 archiveName 'xxxx.war'
}

dependencies {
 compile {redacted}
 compileOnly group: 'javax', name: 'javaee-api', version:'7.0'
 testCompile "junit:junit:4.12"

 compile 'io.swagger:swagger-annotations:1.5.10'
 compile(group: 'com.aaa.inf', name: 'inf-jdbc', version: '3.0.2', classifier: 'sources')
 compile(group: 'com.aaa.inf', name: 'inf-throttle', version: '3.0.1')
 compile(group: 'com.ibm.db2', name: 'db2jcc', version: '3.64.133')
 compile {redacted}
}

jacocoTestCoverageVerification {
 violationRules {
    rule {
        limit {
            minimum = 0.0
        }
    }
 }
}

我的 .ear 的 build.gradle 文件是：

description = 'xxxxx'

buildscript {
repositories {
    maven { url "https://xxxx.aaa.com/artifactory/aaa-mvn" }
}
dependencies {
    classpath "org.jfrog.buildinfo:build-info-extractor-gradle:4.4.5"
    classpath "com.aaa.plugin.gradle:build-defaults-plugin:1.+"
    classpath "com.aaa.plugin.gradle:ucd-publish-plugin:1.+"
    classpath "com.aaa.plugin.gradle:openapi-generator-gradle-plugin:1.+"
}
}

apply plugin: 'ear'
apply plugin: 'maven'
apply from: 'build.local-jboss.gradle'

apply plugin: 'maven-publish'
apply plugin: {redacted} 
apply plugin: 'ucd-publish'
apply plugin: 'build-defaults'
apply plugin: {redacted}
configurations{
    gen
}


dependencies {
   deploy project (path: ":pricingWAR", configuration: 'archives')
   earlib 'com.aaa.inf:inf-api-auth:1.4.+'
   earlib {redacted}

   earlib group: 'javax.security.enterprise', name: 'javax.security.enterprise-api', version: '1.0'
   earlib (group: {redacted}) {
   exclude group: 'javax.security.dddddd'
}
 gen "io.swagger:swagger-codegen:2.2.2"
}

artifacts{
 archives ear
}

repositories {
maven {
    url "https://xxxx.aaa.com/artifactory/aaa-mvn"
}
}

task cleanVolumes(type: Delete) {
delete fileTree(dir: "./volumes/deployments/")
}

task copyEar(type: Copy) {
tasks.cleanVolumes.execute()
from "build/libs"
into "${project.projectDir}/volumes/deployments"
fileMode = 0644
}

build.finalizedBy(generateSwagger)

swaggerConfig {
archive = ear.archivePath
outputFormat = "JSON"
outputPath = project.buildDir.toString() + "/swagger"
}

只是想这样做：将在我的业务类中编译的 .class 文件压缩成 .jar 并将它们添加到 EAR_file\lib。

我什至尝试使用 Gradle 的原生 Groovy 特性来编写构建 .jar 并将其移动到 EAR_file\lib 的代码。这当然是荒谬的。更糟糕的是，它并不完全有效。配合自动部署存在时间问题。

你们中的一些人无疑在畏缩。

为了让我免于这种疯狂，请告诉我你对此的了解。我当然尝试过以下各种排列：

apply plugin: 'java'
...
jar {
  ...
}

无济于事。

提前谢谢你。

Answer 1

在同事的帮助下，终于让它工作了。为了其他人的利益，它最终变成了这样：

在 EAR 项目级别的 build.gradle 中：

 // My project is named: xxx-svc and the .war subproject is named pricing-theAPI
 dependencies {
     deploy project (path: ":xxx-theAPI", configuration: 'archives') // This was here already
     earlib project(path: ":xxx-theAPI", configuration: 'customJar') // This is new
 }

在 WAR 项目级别的 build.gradle 中：

configurations {
  customJar
}

task doJar(dependsOn:classes, type: Jar){
 from sourceSets.main.output 
 include 'com/xyz/xxxxx/yyyyyy/zzzz/**'  // See Note 1
}

artifacts {
  customJar doJar
}

注意 1：我添加了这个修饰符 b/c 在我的情况下，我只需要 .jar 中给定位置的 .class 文件，实际上必须排除所有其他文件。请注意，Gradle 文档错误识别了修饰符“include”。他们称之为“包含”，但实际上它是“包含”。