Tkinter assign button command in loop with lambda(Tkinter 在循环中使用 lambda 分配按钮命令)
问题描述
I'm trying to create a few Button
s (with a for
loop) like so:
def a(self, name):
print(name)
users = {"Test": "127.0.0.0", "Test2": "128.0.0.0"}
row = 1
for name in users:
user_button = Tkinter.Button(self.root, text=name,
command=lambda: self.a(name))
user_button.grid(row=row, column=0)
row += 1
and for the buttons to each get their own parameter (Test getting "Test"
and Test2 getting "Test2"
), but when I press the buttons they both print "Test2"
which means they are using the same function with the same parameter.
How can I solve this?
The problem is your lamba in the for loop. Your lambda is using the name
variable, but the name
variable gets reassigned each time through the for loop. So in the end, all of the buttons get the last value that name
was assigned to in the for loop. To avoid this you can use default keyword parameters in your lamba expression like so:
user_button = Tkinter.Button(self.root,
text=name,
command=lambda name=name: self.a(name))
This binds the current value of the name
variable to the lamba's name
keyword argument each time through the loop, producing the desired effect.
这篇关于Tkinter 在循环中使用 lambda 分配按钮命令的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持编程学习网!
本文标题为:Tkinter 在循环中使用 lambda 分配按钮命令
- 如何在 python3 中将 OrderedDict 转换为常规字典 2022-01-01
- padding='same' 转换为 PyTorch padding=# 2022-01-01
- 分析异常:路径不存在:dbfs:/databricks/python/lib/python3.7/site-packages/sampleFolder/data; 2022-01-01
- python-m http.server 443--使用SSL? 2022-01-01
- 如何将一个类的函数分成多个文件? 2022-01-01
- python check_output 失败,退出状态为 1,但 Popen 适用于相同的命令 2022-01-01
- 沿轴计算直方图 2022-01-01
- 使用Heroku上托管的Selenium登录Instagram时,找不到元素';用户名'; 2022-01-01
- 如何在 Python 的元组列表中对每个元组中的第一个值求和? 2022-01-01
- pytorch 中的自适应池是如何工作的? 2022-07-12