[FIXED] can't scrape the page using request module not working with flask API with BeautifulSoup

Issue

here i'm using flask api module

@app.route('/url', methods=['GET'])
def api_url():
    if 'web_url' in request.args:
        web_url = str(request.args['web_url'])

    html = requests.get(web_url).text
    soup = BeautifulSoup(html, 'lxml')
    web_page = soup.get_text().strip()
    
    return (web_page)

when i give

http://127.0.0.1:5000/url?url=https://stackoverflow.com it's not scraping the webpage , but it's perfectly working with out API

like example

html = requests.get(web_url).text
soup = BeautifulSoup(html, 'lxml')
web_page = soup.get_text().strip()
print(web_page)

and here i just make
import request as requests is that will a issue.? with request.args i just need the scraped web page html code as a output that is i'm searching like when we do on google crome view page source like that way i'm trying to get output

any suggestions

Solution

you should import request from flask and to send the request you, need to import requests

import flask  # to get the args and run the http server
import requests # to send the get request
@app.route('/url', methods=['GET'])
def api_url():
    if 'web_url' in flask.request.args:
        web_url = str(flask.request.args['web_url'])

     html = requests.get(web_url).text
     soup = BeautifulSoup(html, 'lxml')
     web_page = soup.get_text().strip()

     return (web_page)

Answered By - Mangisto