Question

我的应用从其他应用接收文件。所以，我收到了URI但是......如何访问文件名和数据？

现在，我正在做这样的事情：

if ("file".equals(dataUri.getScheme())){
    File file = new File(dataUri.getPath));
    // I do needed operations with the file here
}
else if ("content".equals(dataUri.getScheme())){
    Cursor cursor = getContentResolver().query(dataUri, new String[]{MediaStore.MediaColumns.DISPLAY_NAME}, null, null, null);
    if (cursor.moveToFirst() && (nameIndex = cursor.getColumnIndex(cursor.getColumnNames()[0])) >= 0){
        String fileName = cursor.getString(nameIndex);
        InputStream inputStream = getContentResolver().openInputStream(dataUri);
        // I do needed operations with the file here
    }
}

这足以处理每一个案件吗？

Answer 1

您无需另行处理file和content，可以通过调用getContentResolver().openInputStream(uri)来处理这两种情况。

http://developer.android.com/reference/android/content/ContentResolver.html#openInputStream(android.net.Uri)

从该页面上的其他文档中，您可以看到ContentResolver引用了三个方案，并且该方法可以处理所有这些方案。

SCHEME_ANDROID_RESOURCE
SCHEME_CONTENT
SCHEME_FILE

如果您深入挖掘，可以在此处获取有关打开文档的更多详细信息： http://developer.android.com/guide/topics/providers/document-provider.html#client

虽然它通常是在讨论API 19中提供的更新选项，但以下部分也适用于旧版本。

检查文档元数据

public void dumpImageMetaData(Uri uri) {

    // The query, since it only applies to a single document, will only return
    // one row. There's no need to filter, sort, or select fields, since we want
    // all fields for one document.
    Cursor cursor = getActivity().getContentResolver()
            .query(uri, null, null, null, null, null);

    try {
        // moveToFirst() returns false if the cursor has 0 rows.  Very handy for
        // "if there's anything to look at, look at it" conditionals.
        if (cursor != null && cursor.moveToFirst()) {

            // Note it's called "Display Name".  This is
            // provider-specific, and might not necessarily be the file name.
            String displayName = cursor.getString(
                    cursor.getColumnIndex(OpenableColumns.DISPLAY_NAME));
            Log.i(TAG, "Display Name: " + displayName);

            int sizeIndex = cursor.getColumnIndex(OpenableColumns.SIZE);
            // If the size is unknown, the value stored is null.  But since an
            // int can't be null in Java, the behavior is implementation-specific,
            // which is just a fancy term for "unpredictable".  So as
            // a rule, check if it's null before assigning to an int.  This will
            // happen often:  The storage API allows for remote files, whose
            // size might not be locally known.
            String size = null;
            if (!cursor.isNull(sizeIndex)) {
                // Technically the column stores an int, but cursor.getString()
                // will do the conversion automatically.
                size = cursor.getString(sizeIndex);
            } else {
                size = "Unknown";
            }
            Log.i(TAG, "Size: " + size);
        }
    } finally {
        cursor.close();
    }
}

打开文档

位图

private Bitmap getBitmapFromUri(Uri uri) throws IOException {
    ParcelFileDescriptor parcelFileDescriptor = getContentResolver().openFileDescriptor(uri, "r");
    FileDescriptor fileDescriptor = parcelFileDescriptor.getFileDescriptor();
    Bitmap image = BitmapFactory.decodeFileDescriptor(fileDescriptor);
    parcelFileDescriptor.close();
    return image;
}

输入流

private String readTextFromUri(Uri uri) throws IOException {
    InputStream inputStream = getContentResolver().openInputStream(uri);
    BufferedReader reader = new BufferedReader(new InputStreamReader(inputStream));
    StringBuilder stringBuilder = new StringBuilder();
    String line;
    while ((line = reader.readLine()) != null) {
        stringBuilder.append(line);
    }
    inputStream.close();
    return stringBuilder.toString();
}

如何在我的应用程序中处理收到的文件？

1 个答案: