如何避免烧瓶中的调试模式

Question

请帮助我如何避免 Flask 中的调试模式。它不断给我构建错误

构建错误类似于此

File "C:\Users\HarshitGupta\Desktop\NEw\KoreD\KoreD\templates\flask_user\_common_base.html", line 1, in top-level template code
  {% extends "flask_user/header.html" %}
File "C:\Users\HarshitGupta\Desktop\NEw\KoreD\KoreD\templates\flask_user\header.html", line 60, in top-level template code
  <li><a href="{{ url_for('.home') }}"> <i class="fa fa-pie-chart"></i>Home</a></li>
File "C:\Users\HarshitGupta\Anaconda3\lib\site-packages\flask\helpers.py", line 356, in url_for
  return appctx.app.handle_url_build_error(error, endpoint, values)
File "C:\Users\HarshitGupta\Anaconda3\lib\site-packages\flask\app.py", line 2061, in handle_url_build_error
  reraise(exc_type, exc_value, tb)
File "C:\Users\HarshitGupta\Anaconda3\lib\site-packages\flask\_compat.py", line 35, in reraise
  raise value
File "C:\Users\HarshitGupta\Anaconda3\lib\site-packages\flask\helpers.py", line 345, in url_for
  force_external=external)
File "C:\Users\HarshitGupta\Anaconda3\lib\site-packages\werkzeug\routing.py", line 1776, in build
  raise BuildError(endpoint, values, method, self)
**werkzeug.routing.BuildError: Could not build url for endpoint 'user.home'. Did you mean 'user.login' instead?**

Answer 1

只需使用 app.run(debug=False) 或 app.run()

请检查您是否因为调试模式或其他原因而收到错误！

如何避免烧瓶中的调试模式

How to avoid debugging mode in flask

python

flask