獲取所選文件的完整路徑

Question

我有一個文件夾中的樹狀結構，我想在單擊一個文件時打開該文件。

一切正常，但是當我單擊該文件時，它不會向我發送正確的路徑

import os
import platform
import subprocess
import tkinter as tk
import tkinter.ttk as ttk



class App(tk.Frame):
    def openfolderfile(self,filename2open):
        if platform.system() == "Windows":
            os.startfile(filename2open)
        elif platform.system() == "Darwin":
            subprocess.Popen(["open", filename2open])
        else:
            subprocess.Popen(["xdg-open", filename2open])


    def select(self):
        for i in self.tree.selection():
            print("".join([str(self.tree.item(i)['text'])]))
            print(os.path.abspath("".join([str(self.tree.item(i)['text'])])))
           # self.openfolderfile()

    def __init__(self, master, path):
        tk.Frame.__init__(self, master)
        self.tree = ttk.Treeview(self)
        ysb = ttk.Scrollbar(self, orient='vertical', command=self.tree.yview)
        xsb = ttk.Scrollbar(self, orient='horizontal', command=self.tree.xview)
        self.tree.configure(yscroll=ysb.set, xscroll=xsb.set)
        self.tree.heading('#0', text=path, anchor='w')

        abspath = os.path.abspath(path)
        root_node = self.tree.insert('', 'end', text=abspath, open=True)
        self.process_directory(root_node, abspath)

        self.tree.grid(row=0, column=0,ipady=200,ipadx=200)
        ysb.grid(row=0, column=1, sticky='ns')
        xsb.grid(row=1, column=0, sticky='ew')
        self.grid()
        self.tree.bind("<Double-Button>", lambda e: self.select())




    def process_directory(self, parent, path):
        for p in os.listdir(path):
            abspath = os.path.join(path, p)
            isdir = os.path.isdir(abspath)
            oid = self.tree.insert(parent, 'end', text=p, open=False)
            if isdir:
                self.process_directory(oid, abspath)






root = tk.Tk()
w, h = root.winfo_screenwidth(), root.winfo_screenheight()
root.geometry("%dx%d+0+0" % (w, h))
root.title("Success ")


app = App(root, path="Reports/")
app.mainloop()

這是它打印的內容：

/Users/myuser/PycharmProjects/OCP/Notas.txt

真正的路徑是：

/Users/myuser/PycharmProjects/OCP/Reports/rep1/Notas.txt

為什么會這樣？

Answer 1

像這樣為您的select功能嘗試它：

def select(self):
    for i in self.tree.selection():
        print("".join([str(self.tree.item(i)['text'])]))

        full_path = ''
        current_iid = i

        while True:
            # get next parent directory
            parent_iid = self.tree.parent(current_iid)
            attach_path = self.tree.item(parent_iid)['text']

            # change the parent identifier to be the current now
            current_iid = parent_iid

            # break if the path is empty and thus no more parents are available
            if attach_path == '':
                break

            # add found path to the full path
            full_path = os.path.join(attach_path, full_path)

        print(os.path.join(full_path, ("".join([str(self.tree.item(i)['text'])]))))

您實現的問題是，您使用了os.path.abspath ，這為您提供了當前工作目錄，而不是所選文件所在的目錄。

Answer 2

import os
file = 'myfile.txt'
file = 'directory/to/myfile.txt'
path = os.path.abspath(file)