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Diego
Asked: 2020-11-26 12:39:37 +0800 CST 2020-11-26 12:39:37 +0800 CST

Set a form parameter to a url and get its values ​​in the view

  • 772

I want to be able to search a field for an artist/band from the lastfm api using flask and request. And I have problems in the syntax and/or logic but I don't know how to correct them.

from flask import Flask
from flask import render_template
from flask import request
import requests
import json

app = Flask(__name__)
app.debug = True

@app.route('/')
@app.route('/vista', methods=['GET', 'POST'])
def homepage():
  params = {
    'api_key': '5c1016ab26ab5b2f6ad21c3d873785bc',
    'format' : 'json',
    'artist': request.args.get('artista')
  }
  r = requests.get(
      'http://ws.audioscrobbler.com/2.0/?method=artist.gettopalbums',
      params=params)

  return render_template('vista.html',
               albums=r.json()['topalbums']['album'])


if __name__ == "__main__":
    app.run (host = '127.0.0.1', port = 5000, threaded=True)





   <form action="/buscador.py">
      Artista: <input type="search" name="artista"><br>
      <input type="submit" value="buscar">
    </form>

Debugging

python

1 Answers

  1. Best Answer

    It's not exactly clear what your problem is, but there are several errors in your code:

    1. The api_keymust be a string containing only hexadecimal digits, so the curly braces at the beginning and end are superfluous.
    2. The field artistshould be better within the sent parameters, and not within the URL, because what if the artist name contains spaces or non-ascii characters? It must be properly re-encoded to be a valid URL. requestsit takes care of it if you pass it to it params, but not if you put it in the URL with a simple concatenation.
    3. By default the last.fm API returns the results in XML. Your code seems to assume that the response comes in JSON, so it json.loads()will break. For the response to come in JSON you must add paramsa field formatwith value "json".
    4. The JSON returned, in its field ['topalbums']['album']what it has is a list (it can have 50 elements!). If you want to get just the first one you need to set [0]to select it (and then get its ['name']). It's not clear if your view (template) expects just an album or a list of them, but it ['topalbums']['album']['name']would fail anyway.

    Suppose that the template expects a list of albums and that it itself takes care of accessing the title using an y loop album.nameor similar. This would then be the flask code:

    @app.route('/vista', methods=['GET', 'POST'])
def homepage():
  params = {
    'api_key': '5c1016ab26ab5b2f6ad21c3d873785bc',
    'format' : 'json',
    'artist': request.args.get('artista')
  }
  r = requests.get(
      'http://ws.audioscrobbler.com/2.0/?method=artist.gettopalbums',
      params=params)

  return render_template('vista.html', 
               albums=r.json()['topalbums']['album'])

    Note . To serve your client's request, you're having Flask make another web request. This request requires a time, during which your client will be waiting for the answer. But what's worse, since Flask is single-threaded and single-threaded, the entire server will be blocked until it requestsreturns the result, preventing it from serving more clients. It may be fine for testing, but for production you should deploy Flask via gunicorn, for example, to support concurrency (while a client is waiting for its response, the server can attend to others).

    Update

    The previous code, when used request.args.get('artista')to retrieve the artist to search for from the form, is implicitly expecting that artist to come to it as part of the URL (that is, to come to it in a request like this: GET /vista?artista=Coldplayfor example).

    However, a default HTML form makes a request POSTand encodes the form parameters in a special way as part of the POST body and not in the URL, so the above code will not work.

    An easy way to fix this is to tell the browser that the form should make a request GETinstead of a POST. So:

    <form action="/buscador.py" method="GET">
  Artista: <input type="search" name="artista"><br>
  <input type="submit" value="buscar">
</form>

    Another possibility is to leave the FORM as it was (so it will use POST), and change the Flask part to get the artist's name like this:

      params = {
    ...
    'artist': request.form['artista']
  }

    second update

    Since the user wants to use the same view to show both the FORM and its results (data that I had not understood correctly), it turns out that the same route may or may not carry the parameter artista, depending on whether it has been called as a mere GET /or in response to the submit of the FORM.

    It is necessary to distinguish these cases in the code to avoid the request to the last.fm API if there is no artistasearch.

    The code would be like this:

    @app.route('/')
@app.route('/vista')
def homepage():
  artista = request.args.get('artista')

  # Si no tenemos artista, devolvemos el template que tiene
  # el form y una lista vacía de resultados
  if not artista:
    return render_template('vista.html', albums=[])

  # Si tenemos artista, es como antes
  params = {
    'api_key': '5c1016ab26ab5b2f6ad21c3d873785bc',
    'format' : 'json',
    'artist': artista
  }
  r = requests.get(
      'http://ws.audioscrobbler.com/2.0/?method=artist.gettopalbums',
      params=params)

  return render_template('vista.html', 
               albums=r.json()['topalbums']['album'])

    The FORM must do a GET to the route /vista, so it should be like this:

       <form action="/vista" method="GET">
      Artista: <input type="search" name="artista"><br>
      <input type="submit" value="buscar">
    </form>
    • 2