什么是最简单的方法从android.net.Uri对象持有一个文件:类型转换为java.io.File对象在Android?

我尝试了下面的方法,但不管用:

File file = new File(Environment.getExternalStorageDirectory(), "read.me");
Uri uri = Uri.fromFile(file);
File auxFile = new File(uri.toString());
assertEquals(file.getAbsolutePath(), auxFile.getAbsolutePath());

当前回答

public String getRealPathFromURI(Uri uri) {

    String result;
    Cursor cursor = getContentResolver().query(uri, null, null, null, null);
    if (cursor == null) {
        result = uri.getPath();
        cursor.close();
        return result;
    }
    cursor.moveToFirst();
    int idx = cursor.getColumnIndex(MediaStore.Images.ImageColumns.DATA);
    result = cursor.getString(idx);
    cursor.close();
    return result;
}

然后使用从URI中获取文件:

        File finalFile = newFile(getRealPathFromURI(uri));

——希望能帮到你----

其他回答

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)
}

扩展基于@Jacek kwiecievik回答转换图像uri文件

fun Uri.toImageFile(context: Context): File? {
    val filePathColumn = arrayOf(MediaStore.Images.Media.DATA)
    val cursor = context.contentResolver.query(this, filePathColumn, null, null, null)
    if (cursor != null) {
        if (cursor.moveToFirst()) {
            val columnIndex = cursor.getColumnIndex(filePathColumn[0])
            val filePath = cursor.getString(columnIndex)
            cursor.close()
            return File(filePath)
        }
        cursor.close()
    }
    return null
}

如果我们使用File(uri.getPath()),它将不起作用

如果我们使用扩展从android-ktx,它仍然不能工作,因为 https://github.com/android/android-ktx/blob/master/src/main/java/androidx/core/net/Uri.kt

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。

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

一些库需要文件对象的过程,如改造,一些图像编辑器等。 将URI转换为文件对象或任何类似的东西,我们有一种方法,这是支持所有android旧版本和即将到来的版本。

步骤1:

你必须在FilesDir中创建一个新文件,这个文件对于其他与我们的文件和扩展名相同的应用程序是不可读的。

步骤2:

您必须通过使用InputStream复制URI的内容来创建一个文件。

File f = getFile(getApplicationContext(), uri);

public static File getFile(Context context, Uri uri) throws IOException {
    File destinationFilename = new File(context.getFilesDir().getPath() + File.separatorChar + queryName(context, uri));
    try (InputStream ins = context.getContentResolver().openInputStream(uri)) {
        createFileFromStream(ins, destinationFilename);
    } catch (Exception ex) {
        Log.e("Save File", ex.getMessage());
        ex.printStackTrace();
    }
    return destinationFilename;
}

public static void createFileFromStream(InputStream ins, File destination) {
    try (OutputStream os = new FileOutputStream(destination)) {
        byte[] buffer = new byte[4096];
        int length;
        while ((length = ins.read(buffer)) > 0) {
            os.write(buffer, 0, length);
        }
        os.flush();
    } catch (Exception ex) {
        Log.e("Save File", ex.getMessage());
        ex.printStackTrace();
    }
}

private static String queryName(Context context, Uri uri) {
    Cursor returnCursor =
            context.getContentResolver().query(uri, null, null, null, null);
    assert returnCursor != null;
    int nameIndex = returnCursor.getColumnIndex(OpenableColumns.DISPLAY_NAME);
    returnCursor.moveToFirst();
    String name = returnCursor.getString(nameIndex);
    returnCursor.close();
    return name;
}