使用flavor Dimensions时是否可以排除特定的构建变体？

Question

使用Gradle flavorDimensions时,是否可以排除特定变体？例如 -

android {
...

flavorDimensions "abi", "version"

productFlavors {
    freeapp {
        flavorDimension "version"
        ...
    }

    x86 {
        flavorDimension "abi"
        ...
    }
}

将创建以下构建变体:

x86-freeapp-debug
x86-freeapp-release
arm-freeapp-debug
arm-freeapp-release
mips-freeapp-debug
mips-freeapp-release
x86-paidapp-debug
x86-paidapp-release
arm-paidapp-debug
arm-paidapp-release
mips-paidapp-debug
mips-paidapp-release

可以手动删除"mips-paidapp-release"吗？

Answer 1

从Gradle 0.9开始,您可以应用变量过滤器并对其进行迭代:

productFlavors {
    freeapp {
        dimension "version"
    }

    x86 {
        dimension "abi"
    }

    paidapp {
        dimension "mips"
    }
}

// Loop variants
android.variantFilter { variant ->
    // Loop flavors
    variant.getFlavors().each { flavor ->
        println variant.buildType.name + " " + flavor.name + " " + flavor.dimension
        if (variant.buildType.name.equals('release') &&
                flavor.name.equals('paidapp') &&
                flavor.dimension.equals('mips')) {
            variant.setIgnore(true)
        }
    }
}

注意:我改为flavorDimension,dimension因为后者现在是指定它的首选方式.

注意2:以上注释要求您使用较新的gradle版本:

Project/build.gradle 应该有以下内容:

dependencies {
    classpath 'com.android.tools.build:gradle:1.2.3'
}

虽然Project/app/build.gradle应该有这个:

android {
    buildToolsVersion "22.0.1"
    ...
}