我的脚本和Deform文件上传Widget示例完全一样
@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)使用test_file.grf捕获的上传是一个模式节点,其外观如下:
>> 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
如何将deform.FileData保存为特定位置上的文件?
尝试打开文件并将其复制到src给出的TypeError的位置。
with open(captured['upload']['fp'], 'r') as f:
shutil.copyfileobj(f, src)发布于 2016-06-16 14:13:59
简单地通过二进制打开文件来解决这个问题:
with open(src, 'wb') as f:
f.write(captured['upload']['fp'].read())https://stackoverflow.com/questions/37858285
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