Question

因此我们创建了一个基于文本的冒险游戏的团队项目，并且必须在游戏中加入GUI。虽然GUI和TBA工作正常，但我很难将两者结合起来。我需要用户在Entry Box中输入命令，并根据输入有GUI打印出现有功能中的文本。我正在使用tkinter进行GUI。任何帮助将不胜感激！

基于文本的冒险的一部分

def initWorldVals():
worldVals = [1,0,0,0,0,0,0,'Start Room']
global worldVals

## Room class
class Room:
def __init__(self,x,y,name,desc,memo):
    self.x = x
    self.y = y
    self.name = name
    self.desc = desc
    self.memo = memo

def getX(self):
    return self.x
def getY(self):
    return self.y
def getName(self):
    return self.name
def getDesc(self):
    return self.desc
def getMemo(self):
    return self.memo
numOfRooms = len(roomList)


for i in range(numOfRooms):
  roomList[i] = Room(rooms[i*3],rooms[(i*3)+1],
                   rooms[(i*3)+2],desc[(i*2)+1],memo[i])

 ## Main program functions
 # Room movement

def actionRoom(x,y,commLine,roomList,actionType):
   if commLine == "NORTH":
    y += 1
    if actionType == 'GO':
        for i in range(numOfRooms):
            if roomList[i].getX() == x and roomList[i].getY() == y:
                worldVals[7] = roomList[i].getName()
                worldVals[6] = roomList[i].getY()
        if worldVals[5] != x or worldVals[6] != y:
            errorComm(commLine,'GO')
    elif actionType == 'LOOK':
        for i in range(numOfRooms):
            if roomList[i].getX() == x and roomList[i].getY() == y:
                print(roomList[i].getMemo())
                lookRoomNum = i
                return lookRoomNum
        if roomList[lookRoomNum] != x or roomList[lookRoomNum] != y:
            errorComm(commLine,'LOOK')

elif commLine == "SOUTH":
    y -= 1
    if actionType == 'GO':
        for i in range(numOfRooms):
            if roomList[i].getX() == x and roomList[i].getY() == y:
                if roomList[i] == roomList[5] and worldVals[1] == 0:
                    print("Without looking, you step into the room "
                          "blindly, and fall into a large pit in the "
                          "center of the room.  The fall is too great, "
                          "and you do not survive the fall.\n\n")
                    worldVals[0] = -1
                worldVals[7] = roomList[i].getName()
                worldVals[6] = roomList[i].getY()
        if worldVals[5] != x or worldVals[6] != y:
            errorComm(commLine,'GO')

    elif actionType == 'LOOK':
        for i in range(numOfRooms):
            if roomList[i].getX() == x and roomList[i].getY() == y:
                if roomList[i] == roomList[5]:
                    worldVals[1] = 1
                print(roomList[i].getMemo())
                lookRoomNum = i
                return lookRoomNum
        if roomList[lookRoomNum] != x or roomList[lookRoomNum] != y:
            errorComm(commLine,'LOOK')

GUI代码：

from tkinter import *

import sys

class Display(Frame):

  mGui = Tk()
  mGui.geometry("500x500-500-300") #Size of Widget
  mGui.title("Text Based Adventure!")
  mlabel = Label(mGui,text = "Welcome adventurer where do you wish to go").pack()


def __init__(self):
   Frame.__init__(self)
   self.entry = Entry(self) #Entry textbox
   self.entry.pack()
   self.entry.bind("<Return>", self.onEnter) 
   self.clearButton = Button(self, text="Clear text", command=self.clear_text) #Clear text button
   self.clearButton.pack() 
   self.output = Text(self) #Textbox
   self.output.pack()
   sys.stdout = self
   self.pack()


def onEnter(self,Frame):

    print(self.entry.get()) #Waits for enter to display text


def write(self, txt):
    self.output.insert(END,str(txt)) #displays text



def clear_text(self):
    self.entry.delete(0, "end") #Clears text

if __name__ == '__main__':
  Display().mainloop()

将基于文本的冒险实现到GUI中

0 个答案: