我希望我的应用程序的用户能够删除DCIM文件夹(它位于SD卡上并包含子文件夹)。

这可能吗?如果可能,怎么可能?


当前回答

这是kotlin选项。它运行得很好。

fun executeDelete(context: Context, paths: List<String>): Int {
    return try {
        val files = paths.map { File(it) }
        val fileCommands = files.joinToString(separator = " ") {
            if (it.isDirectory) "'${it.absolutePath}/'" else "'${it.absolutePath}'"
        }
        val command = "rm -rf $fileCommands"
        val process = Runtime.getRuntime().exec(arrayOf("sh", "-c", command))
        val result = process.waitFor()
        if (result == 0) {
            context.rescanPaths(paths)
        }
        result
    } catch (e: Exception) {
        -1
    }
}

//避免在一行中多次调用,它可以删除整个文件夹内容

fun Context.rescanPaths(paths: List<String>, callback: (() -> Unit)? = null) {
if (paths.isEmpty()) {
    callback?.invoke()
    return
}

var cnt = paths.size
MediaScannerConnection.scanFile(applicationContext, paths.toTypedArray(), null) { _, _ ->
    if (--cnt == 0) {
        callback?.invoke()
    }
}
}

其他回答

让我告诉你第一件事你不能删除DCIM文件夹,因为它是一个系统文件夹。当你在手机上手动删除它时,它会删除该文件夹的内容,但不会删除DCIM文件夹。您可以通过以下方法删除其内容:

根据评论更新

File dir = new File(Environment.getExternalStorageDirectory()+"Dir_name_here"); 
if (dir.isDirectory()) 
{
    String[] children = dir.list();
    for (int i = 0; i < children.length; i++)
    {
       new File(dir, children[i]).delete();
    }
}

这是kotlin选项。它运行得很好。

fun executeDelete(context: Context, paths: List<String>): Int {
    return try {
        val files = paths.map { File(it) }
        val fileCommands = files.joinToString(separator = " ") {
            if (it.isDirectory) "'${it.absolutePath}/'" else "'${it.absolutePath}'"
        }
        val command = "rm -rf $fileCommands"
        val process = Runtime.getRuntime().exec(arrayOf("sh", "-c", command))
        val result = process.waitFor()
        if (result == 0) {
            context.rescanPaths(paths)
        }
        result
    } catch (e: Exception) {
        -1
    }
}

//避免在一行中多次调用,它可以删除整个文件夹内容

fun Context.rescanPaths(paths: List<String>, callback: (() -> Unit)? = null) {
if (paths.isEmpty()) {
    callback?.invoke()
    return
}

var cnt = paths.size
MediaScannerConnection.scanFile(applicationContext, paths.toTypedArray(), null) { _, _ ->
    if (--cnt == 0) {
        callback?.invoke()
    }
}
}

如果你不需要递归删除东西,你可以尝试这样做:

File file = new File(context.getExternalFilesDir(null), "");
    if (file != null && file.isDirectory()) {
        File[] files = file.listFiles();
        if(files != null) {
            for(File f : files) {   
                f.delete();
            }
        }
    }

最快最简单的方法:

public static boolean deleteFolder(File removableFolder) {
        File[] files = removableFolder.listFiles();
        if (files != null && files.length > 0) {
            for (File file : files) {
                boolean success;
                if (file.isDirectory())
                    success = deleteFolder(file);
                else success = file.delete();
                if (!success) return false;
            }
        }
        return removableFolder.delete();
}

这是另一种(现代)解决方法。

public class FileUtils {
    public static void delete(File fileOrDirectory) {
        if(fileOrDirectory != null && fileOrDirectory.exists()) {
            if(fileOrDirectory.isDirectory() && fileOrDirectory.listFiles() != null) {      
                Arrays.stream(fileOrDirectory.listFiles())
                      .forEach(FileUtils::delete);
            }
            fileOrDirectory.delete();
        }
    }
}

从API 26开始在Android上运行

public class FileUtils {

    public static void delete(File fileOrDirectory)  {
        if(fileOrDirectory != null) {
            delete(fileOrDirectory.toPath());
        }
    }

    public static void delete(Path path)  {
        try {
            if(Files.exists(path)) {
                Files.walk(path)
                        .sorted(Comparator.reverseOrder())
                        .map(Path::toFile)
//                      .peek(System.out::println)
                        .forEach(File::delete);
            }
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
}