排除特定版本的变种 [英] Exclude specific build variants

问题描述

我有两个默认生成类型：调试/释放和一对夫妇的口味：PROD的/ dev

I have the two default build types: debug / release and a couple of flavors: prod / dev.

现在我要排除的构建变量DEV-释放，但保留其他所有可能的组合。有没有一种方法来实现这一目标？

Now I want to exclude the build variant dev-release, but keep all other possible combinations. Is there a way to achieve this?

推荐答案

由于摇篮插件0.9版本，能够以指定的变体的过滤器。把这样的事情在摇篮构建文件的根目录下：

Variant filter

Since version 0.9 of the gradle plugin, it is possible to specify a variant filter. Put something like this in the root level of the gradle build file:

android.variantFilter { variant ->
  if(variant.buildType.name.equals('release') && variant.getFlavors().get(0).name.equals('vanilla')) {
    variant.setIgnore(true);
  }
}

如果您使用的是多味的设置，您需要做相应的调整。

If you use a multi-flavor setup you'll need to adjust accordingly.

泽维尔的回答，我没有工作，但我想出了这个一个工程，并给出了一个不错的错误信息太多：

Xavier's answer didn't work for me, but I came up with this one that works and gives a nice error message too:

//Disable devRelease because that should not be possible to build
android.applicationVariants.all { variant ->
    if ("devRelease".equals(variant.name)) {
        task assembleDevRelease(overwrite: true) << {
            throw new RuntimeException('Invalid build variant - devRelease should not be built! Did you mean to build prodRelease?');
        }
    }
}

调节devRelease和assembleDevRelease是必要的。

Adjust "devRelease" and "assembleDevRelease" as necessary.

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