android - Gradle buildType/productFlavor 使用意外的 buildConfigField

Question

给定以下配置:

productFlavors {
  normal {
    applicationId "com.app"
  }

  mock {
    applicationId "com.app.mock"
  }
}

buildTypes {
  debug {
    productFlavors.normal.buildConfigField "boolean", "mockMode", "false"
    productFlavors.mock.buildConfigField "boolean", "mockMode", "true"
  }

  release {
    productFlavors.normal.buildConfigField "boolean", "mockMode", "false"
    // Release should never point to mocks. Ever.
    productFlavors.mock.buildConfigField "boolean", "mockMode", "false"
  }
}

我期望 BuildConfig.mockMode = true;，但是，这是生成的构建配置:

public final class BuildConfig {
  public static final boolean DEBUG = Boolean.parseBoolean("true");
  public static final String APPLICATION_ID = "*****";
  public static final String BUILD_TYPE = "debug";
  public static final String FLAVOR = "mock";
  public static final int VERSION_CODE = 1;
  public static final String VERSION_NAME = "1.0";
  // Fields from product flavor: mock
  public static final boolean mockMode = false;
}

通过一些调查/调试，我意识到如果我在发布 buildType 中更改产品风格的值，它实际上会更新 BuildConfig.mockMode 值，尽管有 mockDebug 选择作为我的构建变体。

我已经有了更好的解决方案来实现我想做的事情，所以我只是在寻找一个答案来帮助我理解为什么 Gradle 基于配置以这种方式运行，以帮助我更多地了解它在做什么。

Answer 1

您可以将决定 BuildConfig 字段实际值的逻辑提取到它自己的方法中。这样，DSL 配置只有一行。它看起来像这样(未测试 - 预计语法错误):

buildTypes {
    applicationVariants.all { variant ->
        variant.buildConfigField "boolean", "mockMode", mockMode(variant)
    }
}

def mockMode(variant) {
    //Return true or false depending on variant.buildType and variant.productFlavors 
}