是否可以在 gradle 中创建另一个（构建 APK（s））文件路径？

Question

我希望在项目路径中的不同文件位置生成第二个 debug.apk。是否可以在 gradle 中为 .apk 构建创建第二个文件路径位置？

当前构建路径为：

C:\..\app\build\outputs\apk\debug\debug.apk

我想在构建后创建第二个 apk 位置，例如：

C:\..\app\debug_apk\debug.apk

我目前正在更改 gradle 中输出文件的名称：

applicationVariants.all { variant ->
    variant.outputs.all { output ->
        def appVersionName = "company_name_${versionName}"
        switch (buildType.name) {
            case "debug": {
                outputFileName = "${appVersionName}_debug.apk"
                break
            }
            case "release": {
                outputFileName = "${appVersionName}_release.apk"
                break
            }
        }
    }
}

Answer 1

试试这个解决方案：

---

我可以通过在gradle file 中运行copy 脚本来做到这一点：

    android.applicationVariants.all { variant ->
     variant.outputs.all {

     copy {
      from file("${project.buildDir}/outputs/apk/" + variant.name + "/release/${outputFileName}")
      into file("${project.buildDir}/outputs/apk/")
     }

    // i don't want apk on this location so after successful copy i am deleting it
    delete file("${project.buildDir}/outputs/apk/" + variant.name) 
         }
    }

Answer 2

apply plugin: 'android'

def outputPathName = "D:\build.apk"
def secondDir= "D:\backup\build.apk"

android {
    compileSdkVersion 19
    buildToolsVersion "19.0.3"

    defaultConfig {
        minSdkVersion 8
        targetSdkVersion 19
        versionCode 1
        versionName "1.0"
    }
    buildTypes {
        release {
            runProguard false
            proguardFiles getDefaultProguardFile('proguard-android.txt'), 'proguard-rules.txt'
        }
    }

    def publish = project.tasks.create("publishAll")
    android.applicationVariants.all { variant ->
    variant.outputFile = file(outputPathName)
    def task = project.tasks.create("publish${variant.name}Apk", Copy)
    task.from(variant.outputFile)
    task.into(secondDir)

    task.dependsOn variant.assemble
    publish.dependsOn task
  }

}

dependencies {
    compile 'com.android.support:appcompat-v7:19.+'
    compile fileTree(dir: 'libs', include: ['*.jar'])
}

task cleanExtra(type: Delete) {
    delete outputPathName
}

clean.dependsOn(cleanExtra)

此代码是此resource 和此answer 的组合

Answer 3

这是一个应用模块构建脚本，它提供以下内容：

• 创建 copy*Apk 任务，收集每个构建变体的 APK。
• 创建 copyApks 任务，收集所有构建变体的 APK。
• （可选）使assemble* 任务也收集其特定的 APK。

build.gradle.kts

// Create an umbrella task that will collect APKs from all build variants.
val copyApks = tasks.register("copyApks")

android {
    applicationVariants.all {
        val variantName = name

        // Create variant-specific task that collects the APK.
        val copyApk = tasks.register<Copy>("copy${variantName.capitalize()}Apk") {

            // Copy output files from the task that produces the APK...
            from(packageApplicationProvider)

            // ...into a directory relative to module source root.
            into(file("${variantName}_apk"))

            // Filter out any metadata files, only include APKs.
            include { it.name.endsWith(".apk") }

            // Change the output file name.
            // Only works if there's a single APK for each variant.
            // This will not work with APK splits enabled.
            rename { "${variantName}.apk" }
        }

        // Register the variant-specific task under the umbrella task.
        copyApks.dependsOn(copyApk)

        // (Optional) collect variant-specific APKs when assemble task is invoked.
        assembleProvider.dependsOn(copyApk)
    }
}

build.gradle

如果您使用的是 Groovy，这些行会有所不同：

def copyApks = tasks.register("copyApks")
// ...
def variantName = name
// ...
def copyApk = tasks.register("copy${variantName.capitalize()}Apk", Copy) { // ...

用法

./gradlew copyApks
./gradlew app:copyDebugApk
./gradlew app:copyReleaseApk
./gradlew app:assemble

结果

区别

• 原始 APK 路径未硬编码。相反，它们源自 Android Gradle 插件使用的任何内容。它自动支持product flavors。
• 采集任务依赖原打包任务，参与Gradleup-to-date checks。如果源/APK 未更改，则不会重新收集它们。
• clean 任务不会删除收集的 APK。