Python: 程序在资源管理器中打开了错误的文件路径？

Question

我试图让我的程序打开用户输入的文件路径。此条目保存在我的 data [0] 列表中，然后我使用 os.system 函数调用该列表。资源管理器打开并带我到我的文档，而不是打开保存的文件路径。 1。为什么会这样？ 2.我如何更改此设置以让资源管理器打开保存的用户数据？

import Tkinter as tk
import pickle
import os


data = []

class MyPath(tk.Frame):


def __init__(self, parent):
    tk.Frame.__init__(self, parent)
    self.parent = parent
    self.pack()
    MyPath.make_widgets(self)

def make_widgets(self):
    self.parent.title("MyPath")

    #self.textbox = Text(root)
    #self.textbox.pack()

    # create a prompt, an input box, an output label,
    # and a button to do the computation
    self.prompt = tk.Label(self, text="Enter your Filepath:", anchor="w")
    self.textbox = tk.Text(self)
    self.entry = tk.Entry(self)
    self.open = tk.Button(self, text="Open", command = self.open)
    self.view = tk.Button(self, text="View", command = self.view)
    self.submit = tk.Button(self, text="Save", command = self.save)
    self.output = tk.Label(self, text="")
    self.clear_button = tk.Button(self, text="Clear text", command=self.clear_text)

    # lay the widgets out on the screen. 
    self.prompt.pack(side="top", fill="x")
    self.entry.pack(side="top", fill="x", padx=20)
    self.output.pack(side="top", fill="x", expand=True)
    self.textbox.pack(side="top", fill="x")
    self.view.pack(side="bottom")
    self.submit.pack(side="bottom")
    self.clear_button.pack(side="bottom")
    self.open.pack(side="top")


def clear_text(self):
    self.entry.delete(0, 'end')
    self.textbox.delete(1.0, 'end')


def save(self):
    # get the value from the input widget

    try:
        a = self.entry.get()
        result = data.append(self.entry.get())

    except ValueError:
        result = "Please enter filepaths only"

    #saves list to pickle file
    #pickle.dump(data, open("save.p", "wb"))

    # set the output widget to have our result
    self.output.configure(text=result)
    print (data)

def view(self):
    pickle.load(open("save.p", "rb"))
    self.textbox.insert("1.0", str(pickle.load(open("save.p", "rb"))))
    print pickle.load(open("save.p", "rb"))

def open(self):
    #src = "/Users/matt/Documents/Python/MyApp.py"
    dst = data
    #path = pickle.load(open("save.p", "rb"))
    os.system('Explorer "data[0]"')

if __name__ == "__main__":
    root = tk.Tk()
    MyPath(root).pack(fill="both", expand=True)
    root.mainloop()

非常感谢任何帮助！

Answer 1

os.system('Explorer "%s"' % data[0])

因为您的字符串最后看起来像 'Explorer data[0]'，当资源管理器找不到路径数据时[0]，它会恢复到用户的默认位置 - 我的文档

Python: 程序在资源管理器中打开了错误的文件路径？

Python: Program opens incorrect filepath in explorer?

python

explorer

os.system

list

filepath