我正在尝试将 url 作为 function 中 flask Z23EEEB4947BDD25BDD6

Question

Output This is my code where I'm trying to give url as a parameter in function in flask python but it's showing me this error

url=https://www.amazon.ae/s?i=electronics&rh=n%3A11601326031%2Cn%3A11601327031%2Cn%3A12303750031%2Cn%3A15415001031&s=popularity-rank&pd_rd_r=b187273e-a790-4653-a033-f8e1761d7dcd&pd_rd_w=Zu5vY&pd_rd_wg= bNdaX&pf_rd_p=d90dc948-6b98-4678-a0aa-bc9b5bc35ac8&pf_rd_r=CWGRMPG3B91MY0GE7EPM&ref=pd_gw_unk

@app.route("/data/<url_add>")
def data(url_add):
print(url_add)
headers={'User-Agent': 'Mozilla/5.0 (Macintosh; Intel Mac OS X 10_11_5) AppleWebKit/537.36 (KHTML, like Gecko) Cafari/537.36'}
r = requests.get(url_add, headers=headers)

Answer 1

您正在尝试在path parameters中传递 URL 。 这不方便。 我建议您将其作为论据传递。

这是可能对您有所帮助的代码。

from flask import Flask, request
import requests

app = Flask(__name__)


@app.route("/data", methods=['GET'])
def data():
    url = request.args.get("url")
    print(url)
    headers={'User-Agent': 'Mozilla/5.0 (Macintosh; Intel Mac OS X 10_11_5) AppleWebKit/537.36 (KHTML, like Gecko) Cafari/537.36'}
    r = requests.get(url, headers=headers)
    print(r.text)
    return {"ok": True}

if __name__ == "__main__":
    app.run()

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