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How can i redirect a request with FastAPI if there is a HTTPException?
In Flask we can achieve that like this:
@app.errorhandler(404)
def handle_404(e):
if request.path.startswith('/api'):
return render_template('my_api_404.html'), 404
else:
return redirect(url_for('index'))
Or in Django we can use django.shortcuts:
from django.shortcuts import redirect
def view_404(request, exception=None):
return redirect('/')
How we can achieve that with FastAPI?
935
I know it's too late but this is the shortest approach to handle 404 exceptions in your personal way.
Redirect
from fastapi.responses import RedirectResponse
@app.exception_handler(404)
async def custom_404_handler(_, __):
return RedirectResponse("/")
Custom Jinja Template
from fastapi.templating import Jinja2Templates
from fastapi.staticfiles import StaticFiles
templates = Jinja2Templates(directory="templates")
app.mount("/static", StaticFiles(directory="static"), name="static")
@app.exception_handler(404)
async def custom_404_handler(request, __):
return templates.TemplateResponse("404.html", {"request": request})
Serve HTML from file
@app.exception_handler(404)
async def custom_404_handler(_, __):
return FileResponse('./path/to/404.html')
Serve HTML directly
from fastapi.responses import HTMLResponse
response_404 = """
<!DOCTYPE html>
<html lang="en">
<head>
<title>Not Found</title>
</head>
<body>
<p>The file you requested was not found.</p>
</body>
</html>
"""
@app.exception_handler(404)
async def custom_404_handler(_, __):
return HTMLResponse(response_404)
Note: exception_handler decorator passes the current request and exception as arguments to the function. I've used _ and __ where the variables are unnecessary.
82
answered Dec 05 '25 18:12
We can achieve that by using FastAPI's exception_handler:
If you are in a hurry, you can use this:
from fastapi.responses import RedirectResponse
from starlette.exceptions import HTTPException as StarletteHTTPException
@app.exception_handler(StarletteHTTPException)
async def custom_http_exception_handler(request, exc):
return RedirectResponse("/")
But more spesific approach, you can create your own Exception Handler(s):
class UberSuperHandler(StarletteHTTPException):
pass
def function_for_uber_super_handler(request, exc):
return RedirectResponse("/")
app.add_exception_handler(UberSuperHandler, function_for_uber_super_handler)
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