使用

InputStream inputStream = getContentResolver().openInputStream(uri);    

直接复制文件。还看到:

https://developer.android.com/guide/topics/providers/document-provider.html

编辑:对不起，我之前应该测试得更好。这应该可以工作:

new File(new URI(androidURI.toString()));

URI是java.net.URI。

你可以使用这个函数从uri中获取文件在新的android和旧的

fun getFileFromUri(context: Context, uri: Uri?): File? {
    uri ?: return null
    uri.path ?: return null

    var newUriString = uri.toString()
    newUriString = newUriString.replace(
        "content://com.android.providers.downloads.documents/",
        "content://com.android.providers.media.documents/"
    )
    newUriString = newUriString.replace(
        "/msf%3A", "/image%3A"
    )
    val newUri = Uri.parse(newUriString)

    var realPath = String()
    val databaseUri: Uri
    val selection: String?
    val selectionArgs: Array<String>?
    if (newUri.path?.contains("/document/image:") == true) {
        databaseUri = MediaStore.Images.Media.EXTERNAL_CONTENT_URI
        selection = "_id=?"
        selectionArgs = arrayOf(DocumentsContract.getDocumentId(newUri).split(":")[1])
    } else {
        databaseUri = newUri
        selection = null
        selectionArgs = null
    }
    try {
        val column = "_data"
        val projection = arrayOf(column)
        val cursor = context.contentResolver.query(
            databaseUri,
            projection,
            selection,
            selectionArgs,
            null
        )
        cursor?.let {
            if (it.moveToFirst()) {
                val columnIndex = cursor.getColumnIndexOrThrow(column)
                realPath = cursor.getString(columnIndex)
            }
            cursor.close()
        }
    } catch (e: Exception) {
        Log.i("GetFileUri Exception:", e.message ?: "")
    }
    val path = realPath.ifEmpty {
        when {
            newUri.path?.contains("/document/raw:") == true -> newUri.path?.replace(
                "/document/raw:",
                ""
            )
            newUri.path?.contains("/document/primary:") == true -> newUri.path?.replace(
                "/document/primary:",
                "/storage/emulated/0/"
            )
            else -> return null
        }
    }
    return if (path.isNullOrEmpty()) null else File(path)
}

通过下面的代码，我能够获得adobe应用程序共享pdf文件作为流，并保存到android应用程序路径

Android.Net.Uri fileuri =
    (Android.Net.Uri)Intent.GetParcelableExtra(Intent.ExtraStream);

    fileuri i am getting as {content://com.adobe.reader.fileprovider/root_external/
                                        data/data/com.adobe.reader/files/Downloads/sample.pdf}

    string filePath = fileuri.Path;

   filePath I am gettings as root_external/data/data/com.adobe.reader/files/Download/sample.pdf

      using (var stream = ContentResolver.OpenInputStream(fileuri))
      {
       byte[] fileByteArray = ToByteArray(stream); //only once you can read bytes from stream second time onwards it has zero bytes

       string fileDestinationPath ="<path of your destination> "
       convertByteArrayToPDF(fileByteArray, fileDestinationPath);//here pdf copied to your destination path
       }
     public static byte[] ToByteArray(Stream stream)
        {
            var bytes = new List<byte>();

            int b;
            while ((b = stream.ReadByte()) != -1)
                bytes.Add((byte)b);

            return bytes.ToArray();
        }

      public static string convertByteArrayToPDF(byte[] pdfByteArray, string filePath)
        {

            try
            {
                Java.IO.File data = new Java.IO.File(filePath);
                Java.IO.OutputStream outPut = new Java.IO.FileOutputStream(data);
                outPut.Write(pdfByteArray);
                return data.AbsolutePath;

            }
            catch (System.Exception ex)
            {
                return string.Empty;
            }
        }

安卓 + Kotlin

为Kotlin Android扩展添加依赖项: 实现“androidx.core: core-ktx: {latestVersion}’ 从uri获取文件: uri.toFile ()

你想要的是…

new File(uri.getPath());

．.． ,而不是……

new File(uri.toString());

笔记

对于android.net.Uri对象，它被命名为uri并完全按照问题中创建，uri. tostring()返回一个“file:///mnt/sdcard/myPicture.jpg”格式的字符串，而uri. getpath()返回一个“/mnt/sdcard/myPicture.jpg”格式的字符串。 我知道在Android系统中文件存储有一些细微差别。在这个回答中，我的意图是准确地回答提问者所问的问题，而不是深究其中的细微差别。