为什么我无法使用 VSCode 调试我的烧瓶应用

Question

from   flask  import Flask, render_template, request, url_for, g
app = Flask(__main__)
# This is the entrypoint
@app.route('/')
def index():
# Render page
return 'Hello'
if __name__=='__main__':
app.run(debug=True)

当我尝试在VSCode调试器中运行此程序时，我得到：

* Serving Flask app 'gui' (lazy loading)
* Environment: production
WARNING: This is a development server. Do not use it in a production deployment.
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
No module named gui

我在"；app.run(debug=True("；它确实执行到了这一点，但当我试图越过或进入时，我会出错。

我试着把文件名从gui.py改为app.py，结果得到了同样的结果，只是上面写着：

* Serving Flask app 'app' (lazy loading)
* Environment: production
WARNING: This is a development server. Do not use it in a production deployment.
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
No module named app

我做错了什么/错过了什么？

Answer 1

要使用VSCode调试烧瓶，您需要正确设置setting.json和launch.json文件，这里有一个可能的配置：

settings.json

{
"python.pythonPath": "<your_python_venv_path>",
}

启动.json

{
"version": "0.2.0",
"configurations" : [
{
"name": "local",
"type": "python",
"stopOnEntry": false,
"request": "launch",
"program" : "${workspaceFolder}/<path_to_run.py>",
"console" : "integratedTerminal",
"justMyCode" : false,
"cwd": "${workspaceFolder}"
}
]
}

也可以看看这里