2

我的脚本与变形文件上传小部件示例完全相同:

@view_config(renderer='templates/form.pt', name='file')
@demonstrate('File Upload Widget')
def file(self):

    class Schema(colander.Schema):
        upload = colander.SchemaNode(
            deform.FileData(),
            widget=deform.widget.FileUploadWidget(tmpstore)
            )

    schema = Schema()
    form = deform.Form(schema, buttons=('submit',))

    return self.render_form(form, success=tmpstore.clear)

捕获的上传test_file.grf是一个deform.FileData模式节点,如下所示:

>> captured['upload']
{'filename': u'test_file.grf',
 'fp': <tempfile._TemporaryFileWrapper object at 0x000000000638A6A0>,
  'mimetype': 'text/plain',
  'preview_url': None,
  'size': -1,
  'uid': '42DXY7DYW3'}

问题

如何deform.FileData在特定位置另存为文件?

尝试打开文件并将其复制到src给出的位置TypeError

with open(captured['upload']['fp'], 'r') as f:
    shutil.copyfileobj(f, src)
4

1 回答 1

2

只需通过二进制打开文件即可解决:

with open(src, 'wb') as f:
    f.write(captured['upload']['fp'].read())
于 2016-06-16T14:13:59.200 回答