Query parameters of the get URL using aiohttp from python3.5

Question

async def method(request):
    ## here how to get query parameters
    param1 = request.rel_url.query['name']
    param2 = request.rel_url.query['age']
    return web.Response(text=str(result))


if __name__ == '__main__':
    app = web.Application()
    app.router.add_route('GET', "/sample", method)    

    web.run_app(app,host='localhost', port=3000)

The above code is running in python 3.6. I need to extract the query parameter values (xyz & xy) from the sample URL http://localhost.com/sample?name=xyz&age=xy. I Tried with req.rel_url.query and also with request.query_string. It was giving only the first parameter value xyz, but I was not getting xy (age) which was the second parameter in the query.

How can I get both the query values?

1. localhost was changed to my machine IP , so in query http://<IP Adress>:3000/sample?name=xyz&age=xy 2. business logic for result was not added in the snippet. Since I was failing in request.rel_url.query['age']. — gmrli
– gmrli, Commented Dec 17, 2017 at 1:21

Aleksandar Varicak · Accepted Answer · 2017-12-17 00:46:20Z

16

You have few errors here.

result is not defined in your function. You get params right way, but then error occurs as result is not defined
You're targeting localhost.com, not sure how that is set on your machine, but it shouldn't work.

So, here is working example:

from aiohttp import web

async def method(request):
    ## here how to get query parameters
    param1 = request.rel_url.query['name']
    param2 = request.rel_url.query['age']
    result = "name: {}, age: {}".format(param1, param2)
    return web.Response(text=str(result))


if __name__ == '__main__':
    app = web.Application()
    app.router.add_route('GET', "/sample", method)

    web.run_app(app,host='localhost', port=11111)

then you can try: http://localhost:11111/sample?name=xyz&age=xy and it is working.

answered Dec 17, 2017 at 0:46

Aleksandar Varicak

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5 Comments

gmrli Over a year ago

I tried with the above snippet. I was getting an error of KeyError: "Key not found: 'age'". The business logic for result I have not added. But the issue I was facing was, I was not able to read the second parameter value(In this case it was age )

gmrli Over a year ago

Local host was changed to my machine IP address

Marek Marczak Over a year ago

return web.Response(text=str(result)) str function is not required, because format method returns the string already.

Wayde Over a year ago

Understand this is a old post, but this is exactly what I am trying to do and it works! BUT how do I stop web.run_app once it returns the result? Thanks!

Aleksandar Varicak Over a year ago

Not sure, this was long time ago. You can check documentation to see if it helps: docs.aiohttp.org/en/stable/web_advanced.html#graceful-shutdown

Slipstream · Accepted Answer · 2020-03-21 19:13:50Z

2

Quick example:

async def get_handler(self, request):

    param1 = request.rel_url.query.get('name', '')
    param2 = request.rel_url.query.get('age', '')

    return web.Response(text=f"Name: {param1}, Age: {param2}")

answered Mar 21, 2020 at 19:13

Slipstream

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2 Comments

John Fantastico Feb 7 at 14:43

Needs more detail, you crash your script if the parameter doesn't exist in the above example

Slipstream Feb 27 at 14:11

I understand your concern, but SO answers should address the main question with a very simple example, not a full working program. The developer should be able to understand and change the example accordingly to their needs.

rust · Accepted Answer · 2023-02-03 09:54:00Z

1

The topic is old but I have encountered a similar problem: turns out the issue didn't come from the server but the client:

curl http://localhost.com/sample?name=xyz&age=xy was indeed sending only the first parameter.
However curl "http://localhost.com/sample?name=xyz&age=xy" gave the desired result.

Make sure the parameters are well sent with the request

answered Feb 3, 2023 at 9:54

rust

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