How to fix NameError: name 'html' is not defined?

Question

Python

Hello ! I have this code:

import requests
from bs4 import BeautifulSoup
import pymysql

def get_html(url):
    r = requests.get(url)
    return r.text

# < Сбор ссылок.
def get_links(html):
    soup = BeautifulSoup(html, 'lxml')
    links=soup.findAll(link_container_array[0],{link_container_array[1]:link_container_array[2]})
    #print(resource_name)
    #print(len(links))

    for link_container in links:
        a_tag = link_container .find("a")
        # Если нашел
        if a_tag:
            link = resource_url+ a_tag.get("href")
            #print(link)

# < Сбор контента.
def get_content(html,link):
    soup = BeautifulSoup(html,'lxml')
    headline=soup.find(title_array[0],{title_array[1]:title_array[2]})
    print(headline)

#< Вызывание всех функций.
def get_all_func (resource_allnews_link):
    url = resource_allnews_link
    links = get_links(get_html(url))
    content = get_content(html ,links)

# < Подключение к базе данных.
connection = pymysql.connect(host='localhost',
                             user='root',
                             password='',
                             db='news_portal',
                             charset='utf8',
                             autocommit=True)
cursor = connection.cursor()

# < Запрос правил выдергивания контента.
cursor.execute('SELECT `resource_id`, `resource_name`, `resource_url`, `resource_allnews_link`, `link_rule`, `mainblock`, `mainblock1`, `page_link`, `title`, `datetime`, `datetime1`, `text`, `text1` FROM `resource` WHERE 1')
resources=cursor.fetchall()
# < Цикл для перебора из кортежа.
for resource in resources:
    resource_name=resource[1]
    resource_url=resource[2]
    resource_allnews_link=resource[3]
    content_rule=resource[4]
    title=resource[8]
    resource_name_array=resource_name.split(',')
    link_container_array=content_rule.split(',')
    title_array=title.split(',')
    get_all_func(resource_allnews_link,)

connection.close()

I wanted to make the collected links in the get_links function passed to get_content and after the received data I had the output of my headline
variable. But when I run the script, I get this error

Traceback (most recent call last):
File "C:/Users/Administrator/PycharmProjects/Task/sql_pars.py", line 57, in
get_all_func(resource_allnews_link,)
File "C:/Users/Administrator/PycharmProjects/Task/sql_pars.py". py", line 33, in get_all_func
content = get_content(html ,links)
NameError: name 'html' is not defined
Process finished with exit code 1

PS I understand that my HTML is not defined. I want to know how can I fix this?

Answer 1

Can you just declare this variable?

def get_all_func (resource_allnews_link):
    url = resource_allnews_link
    links = get_links(get_html(url))
    content = get_content(html ,links)

Take a closer look, there are no html variable declarations in this funk.