使用这个来写入文件，它为我工作时，gif的uri是由GBoard提供的，我必须在我的应用程序数据复制该gif。

    try {
      String destinationFilePath = getExternalFilesDir("gifs") + "/tempFile.txt";
      InputStream inputStream = getContentResolver().openInputStream(uri);
      OutputStream outputStream = new FileOutputStream(destinationFilePath);

      byte[] buffer = new byte[1024];
      int bytesRead;

      while ((bytesRead = inputStream.read(buffer)) != -1) {
        outputStream.write(buffer, 0, bytesRead);
      }

      inputStream.close();
      outputStream.close();
    }
    catch (Exception e) {
      e.printStackTrace();
    }

uri.toString()给我:"content://com.google.android.apps.nbu.files.provider/1/file%3A%2F%2F%2Fstorage%2Femulated%2F0%2FDownload%2Fbackup.file"

uri.getPath()给我:“/1/文件:///存储/模拟/0/下载/备份文件。”

new File(uri.getPath())给我“/1/ File:/storage/ emululated /0/Download/backup.file”。

所以如果你有一个文件的访问权限，想要避免使用ContentResolver或直接读取文件，答案是:

private String uriToPath( Uri uri )
{
    File backupFile = new File( uri.getPath() );
    String absolutePath = backupFile.getAbsolutePath();
    return absolutePath.substring( absolutePath.indexOf( ':' ) + 1 );
}

为简化回答，跳过错误处理

最好的解决方案

创建一个简单的FileUtil类，用于创建、复制和重命名文件

我使用了uri.toString()和uri.getPath()，但不适合我。 我终于找到了这个解。

import android.content.Context;
import android.database.Cursor;
import android.net.Uri;
import android.provider.OpenableColumns;
import android.util.Log;

import java.io.File;
import java.io.FileNotFoundException;
import java.io.FileOutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;

public class FileUtil {
    private static final int EOF = -1;
    private static final int DEFAULT_BUFFER_SIZE = 1024 * 4;

    private FileUtil() {

    }

    public static File from(Context context, Uri uri) throws IOException {
        InputStream inputStream = context.getContentResolver().openInputStream(uri);
        String fileName = getFileName(context, uri);
        String[] splitName = splitFileName(fileName);
        File tempFile = File.createTempFile(splitName[0], splitName[1]);
        tempFile = rename(tempFile, fileName);
        tempFile.deleteOnExit();
        FileOutputStream out = null;
        try {
            out = new FileOutputStream(tempFile);
        } catch (FileNotFoundException e) {
            e.printStackTrace();
        }
        if (inputStream != null) {
            copy(inputStream, out);
            inputStream.close();
        }

        if (out != null) {
            out.close();
        }
        return tempFile;
    }

    private static String[] splitFileName(String fileName) {
        String name = fileName;
        String extension = "";
        int i = fileName.lastIndexOf(".");
        if (i != -1) {
            name = fileName.substring(0, i);
            extension = fileName.substring(i);
        }

        return new String[]{name, extension};
    }

    private static String getFileName(Context context, Uri uri) {
        String result = null;
        if (uri.getScheme().equals("content")) {
            Cursor cursor = context.getContentResolver().query(uri, null, null, null, null);
            try {
                if (cursor != null && cursor.moveToFirst()) {
                    result = cursor.getString(cursor.getColumnIndex(OpenableColumns.DISPLAY_NAME));
                }
            } catch (Exception e) {
                e.printStackTrace();
            } finally {
                if (cursor != null) {
                    cursor.close();
                }
            }
        }
        if (result == null) {
            result = uri.getPath();
            int cut = result.lastIndexOf(File.separator);
            if (cut != -1) {
                result = result.substring(cut + 1);
            }
        }
        return result;
    }

    private static File rename(File file, String newName) {
        File newFile = new File(file.getParent(), newName);
        if (!newFile.equals(file)) {
            if (newFile.exists() && newFile.delete()) {
                Log.d("FileUtil", "Delete old " + newName + " file");
            }
            if (file.renameTo(newFile)) {
                Log.d("FileUtil", "Rename file to " + newName);
            }
        }
        return newFile;
    }

    private static long copy(InputStream input, OutputStream output) throws IOException {
        long count = 0;
        int n;
        byte[] buffer = new byte[DEFAULT_BUFFER_SIZE];
        while (EOF != (n = input.read(buffer))) {
            output.write(buffer, 0, n);
            count += n;
        }
        return count;
    }
}

在代码中使用FileUtil类

try {
         File file = FileUtil.from(MainActivity.this,fileUri);
         Log.d("file", "File...:::: uti - "+file .getPath()+" file -" + file + " : " + file .exists());

  } catch (IOException e) {
          e.printStackTrace();
  }

这些对我都没用。我发现这是可行的解决方案。但我的情况仅限于图像。

String[] filePathColumn = { MediaStore.Images.Media.DATA };
Cursor cursor = getActivity().getContentResolver().query(uri, filePathColumn, null, null, null);
cursor.moveToFirst();
int columnIndex = cursor.getColumnIndex(filePathColumn[0]);
String filePath = cursor.getString(columnIndex);
cursor.close();

添加onActivityResult，获取docx或pdf文件

var imageUriPath = ""
imageUriPath =
  if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.O) {
    val split = (imageUri.path ? : "").split(":") //split the path.
    split[1]
  } else {
    imageUri.path ? : ""
  }
val file = File(imageUriPath)

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 ()