从Tkinter Python中的浏览按钮设置Entry值

Question

我不太擅长Python，尤其是在使用类时。我写这段代码是为了使用浏览按钮设置Entry值，问题是这样我应该为每个按钮创建一个浏览方法。有更简单的方法来解决这个问题吗？

from tkinter import *
from tkinter.filedialog import askopenfilename
class App:
def __init__(self, parent):        
self.button1 = Button(text = 'browse', command = self.browse1)     
self.button1.grid (row = 0, column = 1)
self.input_file1 = Entry(textvariable = self.browse1)
self.input_file1.grid(row=0, column = 0)
self.button2 = Button(text = 'browse', command = self.browse2)     
self.button2.grid (row = 1, column = 1)
self.input_file2 = Entry(textvariable = self.browse2)
self.input_file2.grid(row=1, column = 0)
def browse1(self):
filename = askopenfilename(title = 'Select a file')
self.input_file1.delete(0, END)
self.input_file1.insert(0, filename)
def browse2(self):
filename = askopenfilename(title = 'Select a file')
self.input_file2.delete(0, END)
self.input_file2.insert(0, filename)
root = Tk()
root.geometry('900x550')
root.title('prove') 
MyApp = App(root)  
root.mainloop()

谢谢！

Answer 1

如果您的函数如下：

def browse(self, entry):
filename = askopenfilename(title = 'Select a file')
entry.delete(0, END)
entry.insert(0, filename)

然后将您的定义更改为：

self.button1 = Button(text = 'browse', command = lambda: self.browse(self.input_file1))     
self.button1.grid (row = 0, column = 1)
self.input_file1 = Entry()
self.input_file1.grid(row=0, column = 0)

然后，当按下按钮时，它调用lambda函数，该函数调用browse()函数，将相应的输入字段传递给可以插入文本的函数。

希望这是有意义的，如果你有任何问题，请告诉我：(