如何在 Python GUI 中从用户输入的数字中使用任意数量的可用文本框

Posted 2023-02-27

【中文标题】如何在 Python GUI 中从用户输入的数字中使用任意数量的可用文本框

【英文标题】：How to use an arbitrary amount of usable text boxes from a user inputted number in Python GUI

【发布时间】：2013-12-05 06:21:41

【问题描述】：

我正在用 Python 编写一个程序，它是一个骰子游戏的评分系统。游戏可以有任意数量的玩家，所以我有一个输入可以让用户说出他们有多少玩家。

我已经能够将标题标签与每个玩家的名字一起打印到 GUI 中。但是，我尝试为每个人打印一个文本框以便可以输入他们的回合分数，这给我带来了麻烦。我尝试运行一个最多玩家数量的 for 循环，并为每个人打印一个文本框。该方法的问题在于它重复使用了我的self.PlayerXroundScore，因此只有最后一个创建的文本框可用。

这是我的代码，我已尽力注释以使其更易于阅读。

#Allows user to input total number of players
NumPlayers = input("How many players? ")

#Creates a list that is the number of players long
NameList = [0]*NumPlayers

#Allows for input of each Players name
#and stores those names in the list NameList
for i in range(0,NumPlayers):

x = raw_input("Player %d Name? " %(i+1))

NameList[i] = x

#creates the GUI
from Tkinter import *
from tkMessageBox import *

class App(Tk):
   def __init__(self):
       Tk.__init__(self)

       self.Title = ("10,000 scorekeeping")
       self.Header = Label(self, text = "Welcome to 10,000 scoring Module, Have Fun!!", font = ("helvetica", "20", "bold")).grid(row = 0, column = 0, columnspan = (NumPlayers * 3))
       for NameCount in range(1,(NumPlayers+1)):

           #Allows me to create the names as column headers
           self.PlayerName = Label(self, text = "%s" %NameList[NameCount - 1],font = ("helvetica","12","bold")).grid(row = 1, column = ((2 * NameCount)))


           #This if just makes things more aesthetically pleasing, not relevant to my question
           if NameCount < (NumPlayers):

                 self.PlayerName = Label(self, text = "|",font = ("helvetica","12","bold")).grid(row = 1, column = ((2 * NameCount + 1)))


           #This is my problem
           #It succesffully prints the correct number of text boxes
           #however upon button click which calls the vals in each text box
           #only the last created box is useful
           #because that is the box corresponding to PlayerXroundScore
           self.PlayerXroundScore = Entry(self, width = 4)
           self.PlayerXroundScore.grid(row = 2, column = (2 * NameCount))
           self.PlayerXroundScore.insert(0, "0000")

       self.NextRound = Button(self, text = "Next round", command = self.CalcRoundTotals)
       self.NextRound.grid(row = 1, column = 0)

   #This is not completed yet, because I wanted to make sure this is the best way to do it before putting in the time
   #Its obviously doing erroneous things but that will change, 
   #I will encounter the same problem in quite a few different places 
   #but if it can be figured out this once, I can incorporate it elsewhere    
   def CalcRoundTotals(self):

       print x

if __name__ == "__main__":        
    a = App()
    a.mainloop()

这真的让我很烦恼。我考虑过连接，但是，在做self.ConcatenatedVarName = Entry(...) 时我不太知道该怎么做。因为，当我连接时，我会使用eval("Player" + CounterInForLoop + "roundScore")，但 SPE 不喜欢这样做。

任何帮助都会很棒。我真的不想写 50(?) if 语句打印不同数量的文本框if i == NumPlayers

谢谢。

【参考方案1】：

没关系，我自己想出来的，这是其他有类似问题的人的解决方案。

self.PlayerTextBox = []
for NameCount in range(1,(NumPlayers + 1)):
      self.PlayerTextBox.append(Entry(self, width = 4))
      self.PlayerTextBox[NameCount - 1].grid(row = 2, column =   
                                       (2 * NameCount))
      self.PlayerTextBox[NameCount - 1].insert(0, "   0")

就如何打印而言，还有一些其他事情要弄清楚，但这些都是次要的，一旦我尝试，我将能够修复它们。

感谢任何看过它并试图弄清楚的人，即使你没有成功。