如何以及在何处通过基于 macOS Python 的应用程序上的本机 GUI 最好地检索 sudo 密码 - （同时维护交互式输出流（stdout））

Question

好的，情况是这样的：我正在使用 Python 和 wx (wxphoenix) 构建一个 macOS GUI 应用程序。用户可以使用 GUI（例如：script1）启动文件删除过程（包含在 script2 中）。为了运行成功 script2 需要运行具有 sudo 权限。

script2 将遍历一长串文件并将其删除。但是我需要它在每一轮之后与 script1 中包含的 GUI 进行通信，以便 script1 可以更新进度条。

我的当前工作设置绝对是最基本的形式如下所示：

脚本 1:

import io
from threading import Thread
import subprocess

import wx

# a whole lot of wx GUI stuff 

def get_password():
    """Retrieve user password via a GUI"""

    # A wx solution using wx.PasswordEntryDialog()
    # Store password in a variable

    return variable

class run_script_with_sudo(Thread):
    """Launch a script with administrator privileges"""

    def __init__(self, path_to_script, wx_pubsub_sendmessage):
        """Set variables to self"""
        self.path = path_to_script
        self.sender = wx_pubsub_sendmessage
        self.password = get_password()
        Thread.__init__(self)
        self.start()

    def run(self):
        """Run thread"""

        prepare_script = subprocess.Popen(["echo", password], stdout=subprocess.PIPE)
        prepare_script.wait()
        launch_script = subprocess.Popen(['sudo', '-S', '/usr/local/bin/python3.6', '-u', self.path], stdin=prepare_script.stdout, stdout=subprocess.PIPE)
        for line in io.TextIOWrapper(launch_script.stdout, encoding="utf-8"):
            print("Received line: ", line.rstrip())
            # Tell progressbar to add another step:
            wx.CallAfter(self.sender, "update", msg="")

脚本 2：

import time

# This is a test setup, just a very simple loop that produces an output.

for i in range(25):
    time.sleep(1)
    print(i)

以上设置的工作原理是 script1 实时接收 script2 的输出并对其进行操作。（因此在给定的示例中：每秒 script1 向进度条添加另一个步骤，直到达到 25 步）。

我想要实现的目标 = 不将密码存储在变量中并使用 macOS 它的本机 GUI 来检索密码。

然而当我改变时：

prepare_script = subprocess.Popen(["echo", password], stdout=subprocess.PIPE)
prepare_script.wait()
launch_script = subprocess.Popen(['sudo', '-S', '/usr/local/bin/python3.6', '-u', self.path], stdin=prepare_script.stdout, stdout=subprocess.PIPE)
for line in io.TextIOWrapper(launch_script.stdout, encoding="utf-8"):
                print("Received line: ", line.rstrip())
                # Tell progressbar to add another step:
                wx.CallAfter(self.sender, "update", msg="")

进入：

command = r"""/usr/bin/osascript -e 'do shell script "/usr/local/bin/python3.6 -u """ + self.path + """ with prompt "Sart Deletion Process " with administrator privileges'"""
command_list = shlex.split(command)

launch_script = subprocess.Popen(command_list, stdout=subprocess.PIPE)
for line in io.TextIOWrapper(launch_script.stdout, encoding="utf-8"):
    print("Received line: ", line.rstrip())
    # Tell progressbar to add another step:
    wx.CallAfter(self.sender, "update", msg="")

它停止工作，因为显然是 osascript 。这意味着 script2 在完全完成之前不会发送任何输出，导致 script1 中的进度条停止。

我的问题因此变成了：我怎样才能确保使用 macOS 本机 GUI 来询问 sudo 密码，从而避免将其存储在变量中，同时仍然保持在交互式/实时流中从特权脚本捕获标准输出的可能性。

希望这是有道理的。

非常感谢任何见解！

Answer 1

My question thus becomes: How can I make sure to use macOS native GUI to ask for the sudo password, thus preventing having to store it in a variable, while still maintaining the possibility to catch the stdout from the privileged script in an interactive / real-time stream.

我自己找到了解决方案，使用 命名管道 (os.mkfifo())。

这样，您可以让 2 个 python 脚本相互通信，同时其中 1 个通过 osascript 以特权权限启动（意思是：您得到一个本机 GUI window，要求用户 sudo 密码）。

工作解决方案:

mainscript.py

import os
from pathlib import Path
import shlex
import subprocess
import sys
from threading import Thread
import time

class LaunchDeletionProcess(Thread):

    def __init__(self):

        Thread.__init__(self)

    def run(self):

        launch_command = r"""/usr/bin/osascript -e 'do shell script "/usr/local/bin/python3.6 -u /path/to/priviliged_script.py" with prompt "Sart Deletion Process " with administrator privileges'"""
        split_command = shlex.split(launch_command)

        print("Thread 1 started")
        testprogram = subprocess.Popen(split_command)
        testprogram.wait()
        print("Thread1 Finished")

class ReadStatus(Thread):

    def __init__(self):

        Thread.__init__(self)

    def run(self):

        while not os.path.exists(os.path.expanduser("~/p1")):
            time.sleep(0.1)

        print("Thread 2 started")

        self.wfPath = os.path.expanduser("~/p1")

        rp = open(self.wfPath, 'r')
        response = rp.read()

        self.try_pipe(response)

    def try_pipe(self, response):
        rp = open(self.wfPath, 'r')
        response = rp.read()
        print("Receiving response: ", response)
        rp.close()
        if response == str(self.nr_of_steps-1):
            print("Got to end")
            os.remove(os.path.expanduser("~/p1"))
        else:
            time.sleep(1)
            self.try_pipe(response)

if __name__ == "__main__":

    thread1 = LaunchDeletionProcess()
    thread2 = ReadStatus()
    thread1.start()
    thread2.start()

priviliged_script.py

import os
import time
import random

wfPath = os.path.expanduser("~/p1")

try:

    os.mkfifo(wfPath)

except OSError:

    print("error")
    pass

result = 10

nr = 0 

while nr < result:

    random_nr = random.random()

    wp = open(wfPath, 'w')
    print("writing new number: ", random_nr)
    wp.write("Number: " + str(random_nr))       
    wp.close()

    time.sleep(1)
    nr += 1

wp = open(wfPath, 'w')
wp.write("end")     
wp.close()

如何以及在何处通过基于 macOS Python 的应用程序上的本机 GUI 最好地检索 sudo 密码 - （同时维护交互式输出流（stdout））

How & where to best retrieve sudo password via a native GUI on a macOS Python-based app - (while maintaining an interactive output stream (stdout))

python

applescript

subprocess

osascript

python-3.x