Docker error: FileNotFoundError: [Errno 2]

Question

Docker 文件：

FROM python:3
WORKDIR /usr/src/app
RUN pip install flask
RUN pip install flask_restful
RUN pip install jsonpickle
COPY . .
CMD ["python", "restful.py"]

当我在包含 Dockerfile 和 restful.py.

的目录中执行 docker build . 时，它会成功构建

当我执行 docker run e36601b52e23 ls 时，我的输出是：
Dockerfile
restful.py

但是，当我执行 docker run e36601b52e23 时，我的输出是：

j@smith-TDVW1 MINGW64 ~/DockerProjects/python
$ docker run e36601b52e23
 * Serving Flask app "restful" (lazy loading)
 * Environment: production
   WARNING: Do not use the development server in a production environment.
   Use a production WSGI server instead.
 * Debug mode: on
 * Running on http://127.0.0.1:5000/ (Press CTRL+C to quit)
 * Restarting with stat
Traceback (most recent call last):
  File "restful.py", line 76, in <module>
    app.run(debug=True)
  File "/usr/local/lib/python3.7/site-packages/flask/app.py", line 943, in run
    run_simple(host, port, self, **options)
  File "/usr/local/lib/python3.7/site-packages/werkzeug/serving.py", line 988, in run_simple
    run_with_reloader(inner, extra_files, reloader_interval, reloader_type)
  File "/usr/local/lib/python3.7/site-packages/werkzeug/_reloader.py", line 332, in run_with_reloader
    sys.exit(reloader.restart_with_reloader())
  File "/usr/local/lib/python3.7/site-packages/werkzeug/_reloader.py", line 176, in restart_with_reloader
    exit_code = subprocess.call(args, env=new_environ, close_fds=False)
  File "/usr/local/lib/python3.7/subprocess.py", line 323, in call
    with Popen(*popenargs, **kwargs) as p:
  File "/usr/local/lib/python3.7/subprocess.py", line 775, in __init__
    restore_signals, start_new_session)
  File "/usr/local/lib/python3.7/subprocess.py", line 1522, in _execute_child
    raise child_exception_type(errno_num, err_msg, err_filename)
FileNotFoundError: [Errno 2] No such file or directory: '/usr/src/app/restful.py': '/usr/src/app/restful.py'

我正在使用 Windows 10/Git Bash/Docker 桌面。

编辑：我浏览了文件结构，它确实有 /usr/src/app/restful.py。但是，当我执行 python restful.py 命令时，它会抛出该错误。该错误发生在启动命令中。有人知道为什么会这样吗？

#!/usr/bin/python3

from flask import Flask, jsonify, request, abort
from flask_restful import Api, Resource
import jsonpickle

app = Flask(__name__)
api = Api(app)

# Creating an empty dictionary and initializing user id to 0.. will increment every time a person makes a POST request.
# This is bad practice but only using it for the example. Most likely you will be pulling this information from a
# database.
user_dict = {}
user_id = 0


# Define a class and pass it a Resource. These methods require an ID
class User(Resource):
    @staticmethod
    def get(path_user_id):
        if path_user_id not in user_dict:
            abort(400)

        return jsonify(jsonpickle.encode(user_dict.get(path_user_id, "This user does not exist")))

    @staticmethod
    def put(path_user_id):
        update_and_add_user_helper(path_user_id, request.get_json())

    @staticmethod
    def delete(path_user_id):
        user_dict.pop(path_user_id, None)


# Get all users and add new users
class UserList(Resource):
    @staticmethod
    def get():
        return jsonify(jsonpickle.encode(user_dict))

    @staticmethod
    def post():
        global user_id
        user_id = user_id + 1
        update_and_add_user_helper(user_id, request.get_json())


# Since post and put are doing pretty much the same thing, I extracted the logic from both and put it in a separate
# method to follow DRY principles.
def update_and_add_user_helper(u_id, request_payload):
    name = request_payload["name"]
    age = request_payload["age"]
    address = request_payload["address"]
    city = request_payload["city"]
    state = request_payload["state"]
    zip_code = request_payload["zip"]
    user_dict[u_id] = Person(name, age, address, city, state, zip_code)


# Represents a user's information
class Person:
    def __init__(self, name, age, address, city, state, zip_code):
        self.name = name
        self.age = age
        self.address = address
        self.city = city
        self.state = state
        self.zip_code = zip_code


# Add a resource to the api. You need to give the class name and the URI.
api.add_resource(User, "/users/<int:path_user_id>")
api.add_resource(UserList, "/users")

if __name__ == "__main__":
    app.run(debug=True)

第二次编辑：运行 windows 机器上的相同程序：

 * Serving Flask app "app" (lazy loading)
 * Environment: production
   WARNING: Do not use the development server in a production environment.
   Use a production WSGI server instead.
 * Debug mode: on
 * Restarting with stat
 * Debugger is active!
 * Debugger PIN: 511-979-152
 * Running on http://127.0.0.1:5000/ (Press CTRL+C to quit)

当我运行容器在 "Debugger is active!"

之前破裂

Answer 1

所以我找到了一个临时的 "bad" 解决方案。我刚刚从 app.run(debug=True) 中删除了 debug=True，运行没问题。如果有人能解释为什么会这样，那就太好了。

谢谢！

Docker error: FileNotFoundError: [Errno 2]

Docker error: FileNotFoundError: [Errno 2]

python-3.x

docker

dockerfile

docker-desktop