android - 选择构建类型时如何区分构建风格？

Question

我有两个 flavor 维度，staging 和 beta。我有两种口味的苹果和香蕉。 Apple 和 Banana 具有不同的应用使用的 manifestHolder 值。

我的问题是当我要选择一个构建变体时，它结合了两种风格。

即使我为它们分配了不同的维度。

productFlavors {
    apple {
        dimension "staging"
        applicationIdSuffix ".staging"
        manifestPlaceholders = [onesignal_app_id               : "y",
                                // Project number pulled from dashboard, local value is ignored.
                                onesignal_google_project_number: "REMOTE"]
    }

    banana {
        dimension "beta"
        applicationIdSuffix ".beta"
        manifestPlaceholders = [onesignal_app_id               : "x",
                                // Project number pulled from dashboard, local value is ignored.
                                onesignal_google_project_number: "REMOTE"]
    }
}

Answer 1

由于维度，它结合了两者。您只需要一个维度。

假设您有两个 buildTypes:

buildTypes {
    debug {
        debuggable true
        signingConfig signingConfigs.config
    }

    release {
        debuggable false
        minifyEnabled false

        proguardFiles getDefaultProguardFile('proguard-android.txt'), 'proguard-rules.pro'
        signingConfig signingConfigs.config
    }
}

然后为 productFlavors 执行以下操作:

flavorDimensions "default"

productFlavors {
    apple {
        dimension "default"
        applicationIdSuffix ".staging"
        manifestPlaceholders = [onesignal_app_id               : "y",
                                // Project number pulled from dashboard, local value is ignored.
                                onesignal_google_project_number: "REMOTE"]
    }

    banana {
        dimension "default"
        applicationIdSuffix ".beta"
        manifestPlaceholders = [onesignal_app_id               : "x",
                                // Project number pulled from dashboard, local value is ignored.
                                onesignal_google_project_number: "REMOTE"]
    }
}

这将创建 4 个最终构建变体:

苹果调试
appleRelease
香蕉调试
bananaRelease