CODEWITHSUNDEEP
pythonrestfile-transferfastapi
ctenar
ctenar

Reputation: 808

minimalistic way to pass json data to server using FastAPI

I am trying to pass some json contents from the client to the server via some simple REST API built with FastAPI (using uvicorn). If I wrap the file contents into a pydantic.BaseModel like so

app = FastAPI()

class ConfigContents(BaseModel):
    str_contents: str

@app.post("/writeFile/{fn}")
async def write_file(fn: str, contents: ConfigContents):
    filepath = "/some_server_dir/" + fn
    with open(filepath, "w") as f:
        f.write(contents.str_contents)
    return contents

I essentially get what I want, i.e. on the client side (using the request-library), I can execute

response = requests.post("http://127.0.0.1:8000/writeFile/my_file.json", 
             data=json.dumps({"str_contents": contents}))

and end up with the file contents assigned to response and written to the file on the "server". My question is: is there a simpler way to achieve the same, e.g. just passing the json contents as a string to the server without the need to wrap it into a model?

Upvotes: 0

Views: 1508

Answers (1)

Tryph
Tryph

Reputation: 6209

from the fastApi doc:

If you don't want to use Pydantic models, you can also use Body parameters. See the docs for Body - Multiple Parameters: Singular values in body.

The doc for Body explains that you can declare any parameter with a Body(...) default value to make it a value to be retrieved from the resquest body.

This means you could simply delete your pydantic model and change your write_file function declaration to:

async def write_file(fn: str, contents=Body(...)):
   ...

However, it would be a (very) bad idea in my opinion. FastApi uses the pydantic models to validate the provided data and to generate convenient API doc. I would rather recommand to improve and develop the pydantic models you use to get better automatic validation and documentation.

Upvotes: 1

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