CODEWITHSUNDEEP
pythonhttp-status-code-404fastapi
Yagiz Degirmenci
Yagiz Degirmenci

Reputation: 20608

FastAPI handling and redirecting 404

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?

Upvotes: 19

Views: 15272

Answers (4)

javiern8410
javiern8410

Reputation: 23

I find it easier to use Jinja to respond to HTML files. One could use Jinja2 with Jinja2Templates as follows:

templates = Jinja2Templates(directory="./public")

async def not_found(request, exc):
    return templates.TemplateResponse(request, "404.html", status_code=exc.status_code)

exceptions = {
    404: not_found,
}

app = FastAPI(exception_handlers=exceptions)

Next, create a 404.html file in ./public folder.

Upvotes: 1

Shivang Kakkar
Shivang Kakkar

Reputation: 541

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.

Upvotes: 20

Harun Mbaabu Mwenda
Harun Mbaabu Mwenda

Reputation: 19

I uses this method,

from fastapi.responses import RedirectResponse
from starlette.exceptions import HTTPException as StarletteHTTPException

app.mount("/static", StaticFiles(directory="static"), name="static")

templates = Jinja2Templates(directory="templates")

@app.exception_handler(StarletteHTTPException)
async def custom_http_exception_handler(request, exc):
    return templates.TemplateResponse("404.html", {"request": request})
  • Make sure you have static folder, for your static files and templates folder your html files.

Upvotes: 0

Yagiz Degirmenci
Yagiz Degirmenci

Reputation: 20608

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)

Upvotes: 9

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