Question

我使用Python 3.5创建一个简单的GUI，通过textbox从用户接收信息，然后保存到.txt文件。

我注意到的是：

启动时会自动点击SaveButton，但再次点击则不会执行任何操作;
另一个名为tk的空白窗口出现在前景中。

这是我的第一个GUI python应用程序，因此非常感谢简单和耐心。

以下是代码：

#Program by Fares Al Ghazy started 20/5/2017
#Python script to assign key combinations to bash commands, should run in the background at startup
#Since this program is meant to release bash code, it is obviously non-system agnostic and only works linux systems that use BASH
#Further versions which support more OSs may come to life
#This is one file which only creates the GUI, another file is needed to use the info taken by this program

import tkinter as tk
#function to write to file
def SaveFunction(e1,e2):
    print("opening file")
    file = open("BinderData.txt","a")
    combo = e1.get()
    print("combo =" + combo)
    performed = e2.get()
    print("action = " + performed)
    print("Writing to file")
    file.write(combo)
    file.write(performed)
    print("closing file")
    file.close()
    print("file closed")

class MainFrame(tk.Tk):
    def __init__(self,*args,**kwargs):
        tk.Tk.__init__(self,*args,**kwargs)
        #create GUI to take in key combinations and bash codes, then save them in file
        root = tk.Tk() # create new window
        root.wm_title("Key binder") #set title
        #create labels and text boxes
        KeyComboLabel = tk.Label(root,text = "Key combination = ")
        KeyComboEntry = tk.Entry(root)

        ActionLabel = tk.Label(root, text = "Command to be executed = ")
        ActionEntry = tk.Entry(root)
        #place widgets in positions
        KeyComboLabel.grid(row=0,column =0,sticky = tk.E)
        ActionLabel.grid(row=1,column =0, sticky = tk.E)

        KeyComboEntry.grid(row=0,column =1)
        ActionEntry.grid(row=1,column =1)
        #create save button
        SaveButton= tk.Button(root,text = "save")
        SaveButton.grid(row=2,column =2, sticky = tk.E , command = SaveFunction(KeyComboEntry,ActionEntry))

app = MainFrame()
app.mainloop()

Answer 1

执行SaveFunction回调并将结果绑定到command参数;尝试使用lambda表达式。此外，command参数必须转到构造函数，而不是布局函数。
```
SaveButton= tk.Button(root,text = "save", command = lambda: SaveFunction(KeyComboEntry,ActionEntry))
SaveButton.grid(row=2,column =2, sticky = tk.E )
```
您获得了额外的空窗口，因为您创建了两个Tk个实例，第一个是MainFrame本身，扩展了Tk，第二个是Tk()你在__init__创建。不要创建另一个Tk，只需使用self作为root：
```
root = self # or change all "root" to "self" in the code below
```
或者没有MainFrame延长Tk。

有关command回调部分和lambda的更多说明。正如我所说，问题在于您执行该函数，然后将该执行的结果绑定到command。为了更清楚，command=SaveFunction(KeyComboEntry, ActionEntry)与

cmd = SaveFunction(KeyComboEntry, ActionEntry)
SaveButton= tk.Button(root, text="save", command=cmd)

这显然不是你想要的。如果你要调用一个没有参数的函数，你可以只使用command=SaveFunction而没有()，因此不会调用函数，而是使用函数本身作为参数，但是因为{ {1}}需要参数，您必须定义内联SaveFunction，调用lambda，它本身不带参数。

Tkinter创建意外的额外空白窗口

1 个答案: