Question

我对Python很新，但我正在努力学习。我正在使用树莓派，而我正在使用Python 3.2.3版。我的问题是我的框架中有一个文本小部件，可以让用户添加文本。我在下面有一个名为“运行代码”的按钮，当点击它时，它会调用一个函数来检查文本框中的输入。如果字符是按字母顺序排列的，那么它会在标签上打印一条欢迎信息，我的分数应该增加5分，如果输入不是按字母顺序排列则显示错误信息。我试过使用if语句，但只有else部分正在执行。有什么建议？我的代码如下：

    import tkinter as tk
    from tkinter import ttk 





     score = 0


       class Game(tk.Tk): 

        def __init__(self, *args, **kwargs):
        tk.Tk.__init__(self, *args, **kwargs)
        tk.Tk.wm_title(self, "Game") 

        container = tk.Frame(self) 
         container.pack(side="top", fill="both", expand = True)
         container.grid_rowconfigure(0, weight=1) 
         container.grid_columnconfigure(0, weight=1) 


         menubar = tk.Menu(container) 
          filemenu = tk.Menu(menubar, tearoff=0) 
        filemenu.add_command(label="Exit", command=quit)

        menubar.add_cascade(label="File", menu=filemenu) 

        tk.Tk.config(self, menu=menubar)

         self.frames = {}

        for F in (StartPage, Py2Page1): 

             frame = F(container, self) 
             self.frames[F] = frame 

             frame.grid(row=0, column = 0, sticky = "nsew") 

          self.show_frame(StartPage) 
     def show_frame(self, cont): 
         frame = self.frames[cont] 
         frame.tkraise() 


     class StartPage(tk.Frame): 

      def __init__(self, parent, controller):
        tk.Frame.__init__(self,parent) 

        self.columnconfigure(0,weight=1) 
         self.columnconfigure(1,weight=1) 

         label = tk.Label(self, text="Title")
         label.grid(row=0,column=0, columnspan=2)

            label1 = tk.Label(self, text ="Add text here")
         label1.grid(row=1,column=0, columnspan=2, pady=10)


          button1 = ttk.Button(self, text="Start", command=lambda:         controller.show_frame(Py2Page1)) 
            button1.grid(row=2,column=0, columnspan=2,pady=10)





    class Py2Page1(tk.Frame): 
     def __init__(self, parent,controller): 
        tk.Frame.__init__(self, parent)

        result = tk.StringVar()

         self.columnconfigure(0,weight=1,minsize=640) 
             self.columnconfigure(1,weight=1,minsize=640)

         label = tk.Label(self, text="Title!")
         label.grid(row=0,column=0,columnspan=2) 

         self.status = tk.Label(self)
         self.status.grid(row=0,column=1,sticky="e",padx=30) 


         label2 = tk.Label(self, text ="""SubTitle""")
          label2.grid(row=1,column=0)

         label3 = tk.Label(self, text ="""
         Add text here....
         """, justify="left") 
        label3.grid(row=2,column=0,sticky="n")



         self.user_input = tk.Text(self,wrap="word") 
          self.user_input.grid(column=1,row=2)

          button4 = ttk.Button(self,text="Run Code", command = self.checking)
           button4.grid(column=1,row=3)

          button5 = ttk.Button(self,text="Clean", command = self.clear)
           button5.grid(column=1,row=4)

           self.output = tk.Label(self)
           self.output.grid(column=1,row=5,sticky='w', pady=10,padx=10)

        def checking(self):
           answer = self.user_input.get('1.0',tk.END) 

          if  answer.isalpha():
             result = "Hello " + answer
             scorechanged= score =+ 5
          else:
              result = "Please enter your name again"


          self.output.config(text=result)
          self.status.config(text=scorechanged)


      def clear(self):
          self.user_input.delete('1.0',tk.END) 










    app = Game()
     app.geometry("1280x720") 
     app.mainloop()

由于

Answer 1

解决最初的问题：

userInput = stringVar()
userEntry = Entry(master, variable=userInput) #creates an entry box
userInput.get() # returns what is in the entry box

我可以想象用这段代码你可以完成你想要的目标。您可以使用if语句查看输入框中的内容并确定天气是一个数字。

try:
    int(userInput.get()) # try and convert to integer 
except ValueError: # if the conversion fails due to it containing letters
    print('userInput is not a number')

Python tkinter-获取用户输入并检查它

1 个答案: