构建。IS_DEBUGGABLE也可以。它来自于"ro。debuggable"

if (BuildConfig.DEBUG) { 



}

这对我很有用

试试下面的方法:

boolean isDebuggable =  ( 0 != ( getApplicationInfo().flags & ApplicationInfo.FLAG_DEBUGGABLE ) );

科特林：

val isDebuggable = 0 != applicationInfo.flags and ApplicationInfo.FLAG_DEBUGGABLE

这是从这里的邦德尔邮局拍下来的

最简单、最好的长期解决方案是使用BuildConfig.DEBUG。这是一个布尔值，调试版本为真，否则为假:

if (BuildConfig.DEBUG) {
  // do something for a debug build
}

有报道说，这个值在基于eclipse的构建中不是100%可靠的，尽管我个人没有遇到过问题，所以我不能说它到底有多大的问题。

如果你正在使用Android Studio，或者如果你在命令行中使用Gradle，你可以在BuildConfig中添加你自己的东西，或者调整调试和发布构建类型，以帮助在运行时区分这些情况。

The solution from Illegal Argument is based on the value of the android:debuggable flag in the manifest. If that is how you wish to distinguish a "debug" build from a "release" build, then by definition, that's the best solution. However, bear in mind that going forward, the debuggable flag is really an independent concept from what Gradle/Android Studio consider a "debug" build to be. Any build type can elect to set the debuggable flag to whatever value that makes sense for that developer and for that build type.

确保您导入了正确的BuildConfig类 是的，你将没有问题使用:

if (BuildConfig.DEBUG) {
   //It's not a release version.
}

或者，您可以使用BuildConfig.BUILD_TYPE进行区分;

如果您正在运行调试构建 BuildConfig.BUILD_TYPE.equals(“调试”);返回true。对于发布，构建BuildConfig.BUILD_TYPE.equals("release");返回true。