我一直致力于让Gradle根据this blog post生成一个IntelliJ WAR工件定义,作为'idea'任务的一部分。到目前为止,这已经非常成功,虽然我的代码非常脏,所以我正在尝试清理它。 就我而言,war任务应包含在IntelliJ中创建必要标记所需的所有信息。例如,给定一个Gradle Project对象,比如gradleProject,我可以执行以下操作:
def artifactManager = project.component.find { it.@name == 'ArtifactManager' } as Node
if (artifactManager) {
Node artifact = artifactManager.artifact.find { it.@type == 'exploded-war' }
if (artifact) {
artifactManager.remove(artifact)
}
} else {
artifactManager = project.appendNode('component', [name: 'ArtifactManager']);
}
def builder = new NodeBuilder();
def artifact = builder.artifact(type: 'exploded-war', name: "${gradleProject.name} exploded war") {
'output-path'("${gradleProject.buildDir}/${gradleProject.name}-exploded.war")
root(id: 'root') {
element(id: 'directory', name: 'WEB-INF') {
// copy web.xml
element(id: 'file-copy', path: gradleProject.war.webXml)
Set<String> excludeJars = [] as Set
element(id: 'directory', name: 'classes') {
// copy classes
element(id: 'module-output', name: gradleProject.name)
// copy depending projects module output (classes)
gradleProject.configurations.runtime.allDependencies.each {
if (it.hasProperty('dependencyProject')) { // TODO: make it functional?!?
if (it.dependencyProject.hasProperty('jar'))
excludeJars.add(it.dependencyProject.jar.archiveName)
element(id: 'module-output', name: it.dependencyProject.name)
}
}
}
// copy dependency jars excluding possible existing module jars (not sure it's necessary)
element(id: 'directory', name: 'lib') {
// TODO: this should handle ALL kinds of dependencies
gradleProject.configurations.runtime.each {
if (!excludeJars.contains(it.name))
element(id: 'file-copy', path: it)
}
}
}
}
}
artifactManager.append artifact
修改或创建我的IntelliJ .ipr文件。 现在我在Gradle脚本中添加了一些文件到我的WAR工件中,如下所示:
war{
from project(':rest').file('resource/some.properties')
from project(':some_project').fileTree('some_folder')
metaInf {
from parent.file('build/config/website/' + environment + '/context.xml')
}
}
我想添加一些Groovy代码来获取这些文件并创建正确的XML标记,让IntelliJ将它们复制到正确的位置。 最简单的方法是什么?我尝试使用以下内容:
gradleProject.war.source.each { // ... create tags etc. }
但源文件列表包括所有文件,无论其目的地是什么(所有libs,web.xml,类等)。
我如何实现我想要做的事情?
答案 0 :(得分:1)
我终于在Gradle文档和一些试验和错误的帮助下弄明白了。我使用以下代码得到了所需的结果:
def artifactManager = project.component.find { it.@name == 'ArtifactManager' } as Node
if (artifactManager) {
Node artifact = artifactManager.artifact.find { it.@type == 'exploded-war' }
if (artifact) {
artifactManager.remove(artifact)
}
} else {
artifactManager = project.appendNode('component', [name: 'ArtifactManager']);
}
def builder = new NodeBuilder();
def artifact = builder.artifact(type: 'exploded-war', name: "${gradleProject.name} exploded war") {
'output-path'("${gradleProject.buildDir}/${gradleProject.name}-exploded.war")
root(id: 'root') {
println "Adding files to root"
gradleProject.war.rootSpec.children.each {
if (it.destPath.pathString.equals("")) {
add(builder, it, "")
}
}
println "Adding files to META-INF"
element(id: 'directory', name: 'META-INF') {
gradleProject.war.rootSpec.children.each {
if (it.destPath.pathString.equals("META-INF")) {
add(builder, it, "META-INF")
}
}
}
println "Adding files to WEB-INF"
element(id: 'directory', name: 'WEB-INF') {
element(id: 'file-copy', path: gradleProject.war.webXml)
gradleProject.war.rootSpec.children.each {
if (it.destPath.pathString.equals("WEB-INF")) {
add(builder, it, "WEB-INF")
}
}
Set<String> excludeJars = [] as Set
element(id: 'directory', name: 'classes') {
element(id: 'module-output', name: gradleProject.name)
gradleProject.configurations.runtime.allDependencies.each {
if (it.hasProperty('dependencyProject')) {
if (it.dependencyProject.hasProperty('jar'))
excludeJars.add(it.dependencyProject.jar.archiveName)
element(id: 'module-output', name: it.dependencyProject.name)
}
}
}
element(id: 'directory', name: 'lib') {
gradleProject.configurations.runtime.each {
if (!excludeJars.contains(it.name))
element(id: 'file-copy', path: it)
}
}
}
}
}
artifactManager.append artifact
将add
函数定义为:
def add(NodeBuilder builder, Object s, String dest, String prefix = "") {
println(prefix + "current destination: [" + dest + "]")
if (s instanceof File) {
println(prefix + "-- file " + s.toString())
builder.element(id: 'file-copy', path: s)
} else if (s instanceof FileTree) {
println(prefix + "-- folder " + s.dir.toString())
builder.element(id: 'dir-copy', path: s.dir)
} else if (s instanceof CopySpecInternal) {
print(prefix + "-- copy spec " + s.toString() + " with destination path: [" + s.destPath.pathString + "]; ")
if (dest.equals(s.destPath.pathString)) {
println("destination matches current folder")
s.sourcePaths.each { path ->
add(builder, path, dest, prefix + " ")
}
s.children.each { child -> add(builder, child, dest, prefix + " ") }
} else {
println("")
if (s.destPath.segments != null && s.destPath.segments.length > 0 && s.destPath.segments[0].equals(dest)) {
String relDest = new RelativePath(false, Arrays.copyOfRange(s.destPath.segments, 1, s.destPath.segments.length)).pathString
builder.element(id: 'directory', name: relDest) {
s.sourcePaths.each { path ->
add(builder, path, relDest, prefix + " ")
}
s.children.each { child -> add(builder, child, relDest, prefix + " ") }
}
} else {
println(prefix + "!! destination path is not relative to [" + dest + "]: [" + s.destPath.pathString + "]")
}
}
} else println(prefix + "?? unknown " + s.toString())
}
我基本上反向设计了war
任务属性并挖掘到要复制的文件树。据我所知,这个任务没有暴露界面来做这种事情,所以我不得不求助于一个'hacky'解决方案,这不是我的最爱。如果有人能用更清晰的代码填写,我会很高兴。