多个网站上的python 2搜索短语，取自列表文件

Question

So I have following list of links in file called "output":

https://web.archive.org/web/20180101003616/http://onet.pl
https://web.archive.org/web/20180102000139/http://onet.pl
[...]

If you open first link from the list and press "ctrl + f" in firefox, you can find phrase "Katastrofa".

All I want is to have a script, which can find a phrase ("Katastrofa" is only example, I want to use argv argument, but that's not important here), print some success message and proceed further...

I got stuck and can't figure out how to do it. The script I got for testing does not "see" the word ("Katastrofa"), which definitely is on the first page...

Please help :)

Here is what I've done so far:

f = open('output', 'r')
f2 = f.readlines()
for i in f2:
     r=requests.get(i)
     first_page = r.text
     soup = BeautifulSoup(first_page, 'html.parser')
     page_soup = soup
     fraza = "Katastrofa"
     boxes = page_soup.body.find_all(fraza)
     print(i)
     print(boxes)

Output:

https://web.archive.org/web/20180101003616/http://onet.pl

[]
https://web.archive.org/web/20180102000139/http://onet.pl

[]
https://web.archive.org/web/20180103002217/http://onet.pl

Answer 1

if you want to search if in html string contain text

for i in f2:
    r=requests.get(i)
    fraza = "Katastrofa"
    if re.match(fraza, r.text, re.I) # ignore case
        print(i)

if you want to search html element contain text

for i in f2:
    r=requests.get(i)
    soup = BeautifulSoup(r.text, 'html.parser')
    fraza = "Katastrofa"
    boxes = soup.find_all(True, text=re.compile(fraza, re.I))
    if boxes:
        print(i)
        print(boxes)

Results is list of last child element:

https://web.archive.org/web/20180101003616/http://onet.pl
[<span class="title"> Kostaryka: Katastrofa lotnicza. Media: są ofiary  </span>, 
<span class="title"> Australia: katastrofa samolotu, są ofiary śmiertelne  </span>]