Python 获取可执行文件的路径？
Python get path to executable file?
我是 Python 编程的新手，我拼凑了一些代码，希望可以自动重命名特定文件夹中的一堆文件。我需要在没有安装 Python 的计算机上 运行 这个程序（它没有连接到互联网，我无法在上面安装 python）所以我转换了 .py 文件使用 auto-py-to-exe 到 .exe 文件。 .py 文件在我的笔记本电脑上运行，所以只有当我尝试 运行 .exe 文件时才会出现问题。
我在整个代码中使用脚本所在的文件夹的路径，因此在脚本的开头，我将文件路径保存为“路径”：
path = os.path.dirname(os.path.realpath(__file__))
但是，当我将 .py 文件转换为 .exe 文件并将该 .exe 文件移动到另一个文件夹时，保存在 'path' 变量中的文件路径位于临时文件夹中（可能是文件转换器的输出位置）。我通过隔离获取文件路径的脚本部分并制作一个 .exe 文件来仅打印该文件路径来测试它。
有没有其他方法可以获得可执行文件的文件路径？我不想将文件路径硬编码到程序中，因为人们不断地重新组织文件夹那台电脑，我希望这个程序灵活。
如果上下文需要，这里是其余代码。谢谢，如果这是一个非常基本的问题，我很抱歉:)
# This section of code imports the modules used below
import os
import tkinter as tk
from tkinter import simpledialog
from tkinter import messagebox
# Setting global variables
path = os.path.dirname(os.path.realpath(__file__)) + "/" # This sets the filepath to the current folder
# Choosing the dough
dough_input = tk.Tk(className='Select dough')
dough_input.geometry('400x200')
# This is the code that creates the dough selection box. Any new types can be added here using the template below:
dough = tk.StringVar()
radiobutton_1 = tk.Radiobutton(dough_input, text='a', variable=dough, value="a", tristatevalue=0)
radiobutton_1.pack()
radiobutton_2 = tk.Radiobutton(dough_input, text='b', variable=dough, value="b", tristatevalue=0)
radiobutton_2.pack()
radiobutton_3 = tk.Radiobutton(dough_input, text='c', variable=dough, value="c", tristatevalue=0)
radiobutton_3.pack()
radiobutton_4 = tk.Radiobutton(dough_input, text='d', variable=dough, value="d", tristatevalue=0)
radiobutton_4.pack()
# This section of code saves the dough chosen and closes the pop-up window.
def submitfunction():
global bread
bread = dough.get()
dough_input.destroy()
submit = tk.Button(dough_input, text='Submit', command=submitfunction)
submit.pack()
# This section of code triggers the pop-up window for the dough selection.
dough_input.mainloop()
# Inputting the first number
root = tk.Tk()
root.withdraw()
user_inp = simpledialog.askinteger("Number Input", "Input first number:")
i = int(user_inp)
# Defining the file rename function
def rename():
global i # use the variable "i" defined above (The number that the user inputs)
for filename in os.listdir(path): # for each file in the folder specified above do the following
if ".sur" in filename:
my_dest = str(bread) + " " + str(i).zfill(4) + ".txt" # sets the new filename
my_source = path + filename # Defines the old filepath to the file
my_dest = path + my_dest # Defines the new filepath to the file
os.rename(my_source, my_dest) # rename function
i = i + 1 # advances the program down the list
messagebox.showinfo("Success", "All files have been renamed successfully!") # confirmation message!
# This function asks the user if the information they've input is correct. Once the user clicks "ok" the program runs
# the rename function.
def sanity_check():
file_list = os.listdir(path)
j = 0
for g in file_list:
if ".txt" in g:
j += 1
# This section of code does some calculations and creates a "sanity check" for the user. It takes the number that
# the user inputted above and sets that as the first number "i". Then it counts the number of files that
# meets the above criteria and sets that to the number of files. It uses that number of files to calculate the
# last number in the list. It then concatenates all this information and asks the user to
# confirm. If the user presses "ok" then the program
# proceeds with the rename. If the user presses cancel, the program aborts.
first_num = i
num_files = j
length_files = int(num_files) - 1
last_file = i + length_files
last_num = last_file
check = str(
"There are " + str(num_files) + " files in this folder. The dough analyzed is " + str(
dough) + ". The first "
"number is " + str(first_num).zfill(
4) + " and the last number is " + str(
last_num).zfill(4) + ". Is this correct?")
# This is the section of code that asks the user to verify the numbers.
confirmation = messagebox.askokcancel("Sanity Check!", check)
if confirmation == False:
messagebox.showerror("", "Program Cancelled")
else:
rename()
# This is the command that runs the program.
sanity_check()
我想，os.getcwd() 和 sys.executable 就是您要找的。
使用这 2 个函数，您将知道您正在编写 运行 脚本的文件夹（从您启动它的位置），以及当前 python 可执行文件的路径（可能包含在 .exe 分发中, 并解压到 tmp 目录):
$> mkdir -p /tmp/x
$> cd /tmp/x
$> pwd
/tmp/x
$> which python
/usr/bin/python
$> python
Python 3.9.5 (default, May 24 2021, 12:50:35)
[GCC 11.1.0] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> import os
>>> os.getcwd()
'/tmp/x'
>>> import sys
>>> sys.executable
'/usr/bin/python'
>>>
这对我有用。
import sys, os
if getattr(sys, 'frozen', False):
application_path = os.path.dirname(sys.executable)
elif __file__:
application_path = os.path.dirname(__file__)
我也遇到过类似的情况，并且发现无论我是在生产中还是为了测试使用 .exe 文件或 .py 到 运行，上述逻辑都有效。
我是 Python 编程的新手，我拼凑了一些代码，希望可以自动重命名特定文件夹中的一堆文件。我需要在没有安装 Python 的计算机上 运行 这个程序（它没有连接到互联网，我无法在上面安装 python）所以我转换了 .py 文件使用 auto-py-to-exe 到 .exe 文件。 .py 文件在我的笔记本电脑上运行，所以只有当我尝试 运行 .exe 文件时才会出现问题。
我在整个代码中使用脚本所在的文件夹的路径，因此在脚本的开头，我将文件路径保存为“路径”：
path = os.path.dirname(os.path.realpath(__file__))
但是，当我将 .py 文件转换为 .exe 文件并将该 .exe 文件移动到另一个文件夹时，保存在 'path' 变量中的文件路径位于临时文件夹中（可能是文件转换器的输出位置）。我通过隔离获取文件路径的脚本部分并制作一个 .exe 文件来仅打印该文件路径来测试它。
有没有其他方法可以获得可执行文件的文件路径？我不想将文件路径硬编码到程序中，因为人们不断地重新组织文件夹那台电脑，我希望这个程序灵活。
如果上下文需要，这里是其余代码。谢谢，如果这是一个非常基本的问题，我很抱歉:)
# This section of code imports the modules used below
import os
import tkinter as tk
from tkinter import simpledialog
from tkinter import messagebox
# Setting global variables
path = os.path.dirname(os.path.realpath(__file__)) + "/" # This sets the filepath to the current folder
# Choosing the dough
dough_input = tk.Tk(className='Select dough')
dough_input.geometry('400x200')
# This is the code that creates the dough selection box. Any new types can be added here using the template below:
dough = tk.StringVar()
radiobutton_1 = tk.Radiobutton(dough_input, text='a', variable=dough, value="a", tristatevalue=0)
radiobutton_1.pack()
radiobutton_2 = tk.Radiobutton(dough_input, text='b', variable=dough, value="b", tristatevalue=0)
radiobutton_2.pack()
radiobutton_3 = tk.Radiobutton(dough_input, text='c', variable=dough, value="c", tristatevalue=0)
radiobutton_3.pack()
radiobutton_4 = tk.Radiobutton(dough_input, text='d', variable=dough, value="d", tristatevalue=0)
radiobutton_4.pack()
# This section of code saves the dough chosen and closes the pop-up window.
def submitfunction():
global bread
bread = dough.get()
dough_input.destroy()
submit = tk.Button(dough_input, text='Submit', command=submitfunction)
submit.pack()
# This section of code triggers the pop-up window for the dough selection.
dough_input.mainloop()
# Inputting the first number
root = tk.Tk()
root.withdraw()
user_inp = simpledialog.askinteger("Number Input", "Input first number:")
i = int(user_inp)
# Defining the file rename function
def rename():
global i # use the variable "i" defined above (The number that the user inputs)
for filename in os.listdir(path): # for each file in the folder specified above do the following
if ".sur" in filename:
my_dest = str(bread) + " " + str(i).zfill(4) + ".txt" # sets the new filename
my_source = path + filename # Defines the old filepath to the file
my_dest = path + my_dest # Defines the new filepath to the file
os.rename(my_source, my_dest) # rename function
i = i + 1 # advances the program down the list
messagebox.showinfo("Success", "All files have been renamed successfully!") # confirmation message!
# This function asks the user if the information they've input is correct. Once the user clicks "ok" the program runs
# the rename function.
def sanity_check():
file_list = os.listdir(path)
j = 0
for g in file_list:
if ".txt" in g:
j += 1
# This section of code does some calculations and creates a "sanity check" for the user. It takes the number that
# the user inputted above and sets that as the first number "i". Then it counts the number of files that
# meets the above criteria and sets that to the number of files. It uses that number of files to calculate the
# last number in the list. It then concatenates all this information and asks the user to
# confirm. If the user presses "ok" then the program
# proceeds with the rename. If the user presses cancel, the program aborts.
first_num = i
num_files = j
length_files = int(num_files) - 1
last_file = i + length_files
last_num = last_file
check = str(
"There are " + str(num_files) + " files in this folder. The dough analyzed is " + str(
dough) + ". The first "
"number is " + str(first_num).zfill(
4) + " and the last number is " + str(
last_num).zfill(4) + ". Is this correct?")
# This is the section of code that asks the user to verify the numbers.
confirmation = messagebox.askokcancel("Sanity Check!", check)
if confirmation == False:
messagebox.showerror("", "Program Cancelled")
else:
rename()
# This is the command that runs the program.
sanity_check()
我想，os.getcwd() 和 sys.executable 就是您要找的。
使用这 2 个函数，您将知道您正在编写 运行 脚本的文件夹（从您启动它的位置），以及当前 python 可执行文件的路径（可能包含在 .exe 分发中, 并解压到 tmp 目录):
$> mkdir -p /tmp/x
$> cd /tmp/x
$> pwd
/tmp/x
$> which python
/usr/bin/python
$> python
Python 3.9.5 (default, May 24 2021, 12:50:35)
[GCC 11.1.0] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> import os
>>> os.getcwd()
'/tmp/x'
>>> import sys
>>> sys.executable
'/usr/bin/python'
>>>
这对我有用。
import sys, os
if getattr(sys, 'frozen', False):
application_path = os.path.dirname(sys.executable)
elif __file__:
application_path = os.path.dirname(__file__)
我也遇到过类似的情况，并且发现无论我是在生产中还是为了测试使用 .exe 文件或 .py 到 运行，上述逻辑都有效。