如何将文件的路径发布到tkinter中的entry()中
如何将文件的路径发布到tkinter中的entry()中
提问于 2019-05-26 17:24:22
我正在尝试创建(text1)条目来获取我打开的文件路径以显示我的意思当我打开一个文件时它会在条目(文本框)上显示文件路径ps:很抱歉我的例子和英语都不好
‘python
import tkinter as tk
from tkinter.filedialog import askopenfilename
from tkinter.messagebox import showerror
from tkinter import ttk这是我的fream课程
class SchoolProjict(tk.Tk):
def __init__(self, *args, **kwargs):
tk.Tk.__init__(self, *args, **kwargs)
container = tk.Frame(self)
container.pack(side = "top", fill = "both", expand = True)
container.grid_rowconfigure(0, weight = 1)
container.grid_columnconfigure(0, weight = 1)
self.frames = {}
for F in (StartPage, PageOne, SetingPage):
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()此选项用于测试从条目打印
def printingstuff(var1):
print (var1)这是为了打开一个文件，我想让它更改条目以显示文件路径
def load_file():
fname = askopenfilename(filetypes=(("Excel file", "*.xls"),
("HTML files", "*.html;*.htm"),
("All files", "*.*") ))
if fname:
try:
print(fname)
return
except: # <- naked except is a bad idea
showerror("Open Source File", "Failed to read file\n'%s'" % fname)
return以下是程序的框架
class StartPage(tk.Frame):
def __init__(self, parent, controller):
tk.Frame.__init__(self, parent)
lablel = tk.Label(self, text = "Start Page")
lablel.pack(pady = 10, padx = 10)
button1 = tk.Button(self, text = "Main Menu", command = lambda: controller.show_frame(PageOne))
button1.pack()
button2 = tk.Button(self, text = "Siting", command = lambda: controller.show_frame(SetingPage))
button2.pack()
class PageOne(tk.Frame):
def __init__(self, parent, controller):
tk.Frame.__init__(self, parent)
lablel = tk.Label(self, text = "Main Menu")
lablel.pack(pady = 10, padx = 10)
button1 = tk.Button(self, text = "Start Page", command = lambda: controller.show_frame(StartPage))
button1.pack()
button2 = tk.Button(self, text = "Siting", command = lambda: controller.show_frame(SetingPage))
button2.pack()
class SetingPage(tk.Frame):
def __init__(self, parent, controller):
tk.Frame.__init__(self, parent)
lablel = tk.Label(self, text = "Siting Page")
lablel.pack(pady = 10, padx = 10)
text1 = ttk.Entry(self)
text1.pack()
text1.focus()
button1 = tk.Button(self, text = "Print from Entry", command = lambda: printingstuff(text1.get()))
button1.pack()
button2 = tk.Button(self, text="open File", command= load_file, width=10)
button2.pack()
button3 = tk.Button(self, text = "Main Menu", command = lambda: controller.show_frame(PageOne))
button3.pack()
button4 = tk.Button(self, text = "Start Page", command = lambda: controller.show_frame(StartPage))
button4.pack()主循环的事情
app = SchoolProjict()
app.mainloop()‘如果这没有任何意义，我很抱歉
回答 1
Stack Overflow用户
回答已采纳
发布于 2019-06-04 01:09:14
class SetingPage(tk.Frame):
def __init__(self, parent, controller):
...
self.text1 = tk.Entry(self) #<== i want to show the path of the file i am going to open Here after i select it from openfile
self.text1.grid(row = 2, column = 0)
self.text1.focus()
button1 = tk.Button(self, text = "print text1", command = lambda: printingstuff(self.text1.get()))
...页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/56315826
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