list - 如何从非流函数获取流输出

Question

我想要来自非流函数(返回类型 Flow<T> )的流输出(返回类型 T )。

fun getTotalFiles(): Int 
// Say, This is a library function it'll return the number of files (Int) in that folder at that specific moment.

//And,
fun getAllFiles(): List<File> 
// Say, This is a library function it'll return all the files (List<File>) in that folder.

该文件夹中的文件将来可能并且将会更改。

现在，我想不断观察输出，那么我该如何实现呢？

fun getFlowOfTotalFiles(): Flow<Int> =
// A wrapper function that converts the library function return type to an observable flow, Flow<Int>

//And,
fun getFlowOfAllFiles(): Flow<List<File>> =
// A wrapper function that converts the library function return type to an observable flow, Flow<List<File>>

Answer 1

为了专门监视目录中的文件，您可以使用 WatchService 并使用 flow 构建器将其转换为流。像这样的事情:

fun getDirectoryMonitorFlow(directory: String) = flow {
    FileSystems.getDefault().newWatchService().use { watchService ->
        while (true) {
            val watchKey = Path.of(directory).register(watchService, ENTRY_CREATE, ENTRY_DELETE, ENTRY_MODIFY)
            if (watchKey.pollEvents().isNotEmpty()) {
                emit(Unit)
            }
            yield() // give flow opportunity to be cancelled.
            if (!watchKey.reset()) {
                println("Directory became unreadable. Finishing flow.")
                break
            }
        }
    }
}
    .catch { println("Exception while monitoring directory.") }
    .flowOn(Dispatchers.IO)

然后你的类可能看起来像:

fun getFlowOfTotalFiles(): Flow<Int> = getFlowOfAllFiles()
    .map { it.size }
    .distinctUntilChanged()

fun getFlowOfAllFiles(): Flow<List<File>> = flow {
    emit(Unit) // so current state is always emitted
    emitAll(getDirectoryMonitorFlow(directory))
}
    .map {
        File(directory).listFiles()?.toList().orEmpty()
    }
    .flowOn(Dispatchers.IO)
    .distinctUntilChanged()

尽管您可能会考虑将第一个流设为私有(private) SharedFlow，这样您就不会运行多个 WatchService 来同时监视同一目录。