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code/flask-debug-run
Flask app run with debug=True
What it detects
`app.run(debug=True)` enables the Werkzeug debugger, which exposes an interactive Python console to anyone who can trigger an exception — remote code execution if reachable.
How it runs
Gated on framework detection: the repo's manifests (package.json, requirements.txt, pom.xml, Gemfile, composer.json) are read to identify the stack, and the rule only runs against matching-language files when its framework is present.
Found a false positive or want this rule tuned? File an issue. You can also suppress per-repo via a .repoguardignore line.