Question

我正在使用this question的答案并看到评论：

   raw = data.file.read() # This is dangerous for big files

如何在不执行此操作的情况下上传文件？到目前为止我的代码是：

@bottle.route('/uploadLO', method='POST')
def upload_lo():
    upload_dir = get_upload_dir_path()
    files = bottle.request.files
    print files, type(files)
    if(files is not None):
        file = files.file
        print file.filename, type(file)
        target_path = get_next_file_name(os.path.join(upload_dir, file.filename))
        print target_path
        shutil.copy2(file.read(), target_path)  #does not work. Tried it as a replacement for php's move_uploaded_file
    return None

给出了这个输出：

127.0.0.1 - - [03/Apr/2014 09:29:37] "POST /uploadLO HTTP/1.1" 500 1418
Traceback (most recent call last):
  File "C:\Python27\lib\site-packages\bottle.py", line 862, in _handle
    return route.call(**args)
  File "C:\Python27\lib\site-packages\bottle.py", line 1727, in wrapper
    rv = callback(*a, **ka)
  File "C:\dev\project\src\mappings.py", line 83, in upload_lo
    shutil.copy2(file.read(), target_path)
AttributeError: 'FileUpload' object has no attribute 'read'

我正在使用python bottle v.12，dropzone.min.js和mongodb。我也在使用本教程：

http://www.startutorial.com/articles/view/how-to-build-a-file-upload-form-using-dropzonejs-and-php

Answer 1

这称为“文件啜食”：

raw = data.file.read()

和you don't want to do it（至少在这种情况下）。

这是一种更好的方法来读取未知（可能很大）的二进制文件：

data_blocks = []

buf = data.file.read(8192)
while buf:
    data_blocks.append(buf)
    buf = data.file.read(8192)

data = ''.join(data_blocks)

如果累积的大小超过某个阈值，您可能还想停止迭代。

希望有所帮助！

第2部分

你问过限制文件大小，所以这是修改后的版本：

MAX_SIZE = 10 * 1024 * 1024 # 10MB
BUF_SIZE = 8192

# code assumes that MAX_SIZE >= BUF_SIZE

data_blocks = []
byte_count = 0

buf = f.read(BUF_SIZE)
while buf:
    byte_count += len(buf)

    if byte_count > MAX_SIZE:
        # if you want to just truncate at (approximately) MAX_SIZE bytes:
        break
        # or, if you want to abort the call
        raise bottle.HTTPError(413, 'Request entity too large (max: {} bytes)'.format(MAX_SIZE))

    data_blocks.append(buf)
    buf = f.read(BUF_SIZE)

data = ''.join(data_blocks)

它并不完美，但它很简单，IMO足够好。

Answer 2

要添加到ron.rothman的优秀答案......修复错误消息，请尝试以下

@bottle.route('/uploadLO', method='POST')
def upload_lo():
    upload_dir = get_upload_dir_path()
    files = bottle.request.files

    # add this line
    data = request.files.data

    print files, type(files)

    if(files is not None):
        file = files.file
        print file.filename, type(file)
        target_path = get_next_file_name(os.path.join(upload_dir, file.filename))
        print target_path

        # add Ron.Rothman's code
        data_blocks = []
        buf = data.file.read(8192)
        while buf:
            data_blocks.append(buf)
            buf = data.file.read(8192)

        my_file_data = ''.join(data_blocks)
        # do something with the file data, like write it to target
        file(target_path,'wb').write(my_file_data)

    return None

Python瓶 - 如何在不管理服务器的情况下上传媒体文件

2 个答案: