使用 ipywidgets FileUpload 小部件时,我可以定义文件上传操作吗
Can I define an action on file upload when using ipywidgets FileUpload widget
我想在使用 widgets.FileUpload
上传文件时创建一个事件,就像我使用 .on_click(some_function)
对 widgets.Button
所做的一样。有没有办法做这样的事情
import ipywidgets as widgets
def do_something_with_file(fu):
content = fu.data[0].decode('utf-8')
# doing something with the file content and get some string output
return 'some string'
str_file_upload = widgets.FileUpload(accept='.txt', multiple=False)
str_file_upload.on_upload(do_something_with_file)
可以观察_counter的变化
示例:
from ipywidgets import widgets
uploader = widgets.FileUpload()
display(uploader)
def on_upload_change(change):
print(change)
uploader.observe(on_upload_change, names='_counter')
我想在使用 widgets.FileUpload
上传文件时创建一个事件,就像我使用 .on_click(some_function)
对 widgets.Button
所做的一样。有没有办法做这样的事情
import ipywidgets as widgets
def do_something_with_file(fu):
content = fu.data[0].decode('utf-8')
# doing something with the file content and get some string output
return 'some string'
str_file_upload = widgets.FileUpload(accept='.txt', multiple=False)
str_file_upload.on_upload(do_something_with_file)
可以观察_counter的变化
示例:
from ipywidgets import widgets
uploader = widgets.FileUpload()
display(uploader)
def on_upload_change(change):
print(change)
uploader.observe(on_upload_change, names='_counter')