Python 3.x Python使用变量打开链接

我用python创建的代码有问题。 我想URL API电报打开一个变化，以便从网站下载的项目被发送到聊天室

# Import libraries
import requests
import urllib.request
import time
import sys 
from bs4 import BeautifulSoup
stdoutOrigin=sys.stdout 
sys.stdout = open("log.txt", "w")
# Set the URL you want to webscrape from
url = 'https://31asdasdasdasdasd.com/'
# Connect to the URL
response = requests.get(url)
# Parse HTML and save to BeautifulSoup object
soup = BeautifulSoup(response.text, "html.parser")
zapisane = ''
row = soup.find('strong')
print(">> Ilosc opinii ktora przeszla:")
send =  print(row.get_text()) # Print row as text
import urllib.request
u = urllib.request.urlopen("https://api.telegram.org/botid:ts/sendMessage?chat_id=-3channel1&text=")

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您可能希望在此处显示的最后一行代码中使用带有变量的字符串格式。以下是有关字符串格式设置的有用资源：