如何使用pymongo从mongodb获取文件对象？

Jeff 发表于 Dev

杰夫

我目前正在使用：

some_fs = gridfs.GridFS(db, "some.col")
fs_file = some_fs.get(index)

得到一个<class 'gridfs.grid_file.GridOut'>对象。

如何获取文件对象，或者如何将其转换为python文件对象？为此，我是否需要另存为临时文件？

编辑：

这是我正在使用的完整代码：

FFMPEG_BIN = "ffmpeg.exe"
some_fs = gridfs.GridFS(db, "some.col")
vid_id = ObjectId("5339e3b5b322631b544b2338")

vid_file = some_fs.get(vid_id)
raw =  vid_file.read()
print type(vid_file), type(raw)

with open(raw, "rb") as infile:
    pipe = sp.Popen([FFMPEG_BIN,
                 # "-v", "quiet",
                 "-y",
                 "-i", "-",
                 "-vcodec", "copy", "-acodec", "copy",
                 "-ss", "00:00:00", "-t", "00:00:10", "-sn",
                 "test.mp4" ]
                ,stdin=infile, stdout=sp.PIPE
)
pipe.wait()

输出：

[2014-03-31 19:03:00] Connected to DB.
<class 'gridfs.grid_file.GridOut'> <type 'str'>
Traceback (most recent call last):
  File "C:/dev/proj/src/lib/ffmpeg/win/test.py", line 19, in <module>
    with open(raw, "rb") as infile:
TypeError: file() argument 1 must be encoded string without NULL bytes, not str

尼克·伦佩尔

编辑：也许GridOut不是python文件对象的正确实现。我的最后建议是尝试将内存文件与StringIO一起使用。

import StringIO

FFMPEG_BIN = "ffmpeg.exe"
some_fs = gridfs.GridFS(db, "some.col")
vid_id = ObjectId("5339e3b5b322631b544b2338")

vid_file = some_fs.get(vid_id)

# Should be a proper file-like object
infile =  StringIO.StringIO(vid_file.read())

pipe = sp.Popen([FFMPEG_BIN,
    # "-v", "quiet",
    "-y",
    "-i", "-",
    "-vcodec", "copy", "-acodec", "copy",
    "-ss", "00:00:00", "-t", "00:00:10", "-sn",
    "test.mp4" ]
    ,stdin=infile, stdout=sp.PIPE
)
pipe.wait()

...

infile.close()

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