将此Python代码转换为可视GUI?

时间:2013-04-06 19:40:33

标签: python user-interface

我有一个朋友制作的python代码,因为我对python知之甚少。 我想将它转换为GUI,因为我们将代码作为程序分发,以便初学者友好。

无论如何这里是代码:

import time, os, sys 
try: 
if len(sys.argv) < 2: 
    fn = raw_input("Enter the name of the file you want to edit: ") 
else: fn = sys.argv[1] 
f = open(fn) 
b = f.read() 
for i in b[:300]: 
    print hex(ord(i))[2:], 
f.close() 
line = str(0x15c)+'-'+str(0x15f) 
if len(sys.argv)<3: 
    hexcode = raw_input("3 bytes color hex number: ") 
else: hexcode = sys.argv[2] 
if not hexcode.startswith('0x'): 
    hexcode = '0x'+hexcode 
hexstr = '0x'
start = int(line.split('-')[0]) 
end = int(line.split('-')[1]) 
for i in b[start:end]: 
    hexstr+=hex(ord(i))[2:] 
ascii = '' 
for i in range(2,len(hexcode),2): 
    char = chr(int(hexcode[i:i+2],16)) 
    ascii+=char 
b = b[:start]+ascii+b[end:] 
for i in b[:300]: 
    print hex(ord(i))[2:], 
except Exception, x: 
print x 
time.sleep(3) 
finally: 
f = open(fn,'wb') 
f.write(b) 
f.close() 

现在我在教程中发现了这个,但不知道如何使用它:     #simple GU0I

from Tkinter import *

root = Tk()
root.title("BreffHexReplace")
root.geometry("400x200")

app = Frame(root)
app.grid()
label = Label(app, text = "This is a label!")

label.grid()

root.mainloop()

有任何帮助吗? 谢谢!

还有一件事,在这段代码中,输入名称或者替换十六进制后,它会显示十六进制列表,如何才能显示?

谢谢!

1 个答案:

答案 0 :(得分:0)

因为这个程序很简单,你可以试试EasyGui(easygui.sourceforge.net/)。首先,添加import easygui as eg。该行加载EasyGui模块。接下来,将fn = raw_input("Enter the name of the file you want to edit: ")替换为fn = eg.fileopenbox(title = 'HexReplace', msg = 'Browse to the file you wish to edit')。这将打开一个框以浏览到所需文件。同时将hexcode = raw_input("3 bytes color hex number: ")替换为hexcode = eg.enterbox(msg = '3 bytes color hex number', title = 'HexReplace')。这会弹出一个输入框。将print x替换为eg.msgbox(title = 'HexReplace', msg = x)。这显示了一个带有异常的消息框。 title参数是窗口标题。要删除十六进制列表,请在导入和try块之间添加:null = open(os.devnull, 'W's); oldstdout = system.stdout; sys.stdout = null 这将使所有打印消息静音。