通过将其与列表（Python）进行比较来从网页中删除项目

Question

I have gathered the data in a list which needs to be removed , the below code shows the list :

keyword= "www.indigo.com"
hrefs = [links['href'] for links in getDetails.find_all('a', href=True) if target in links['href']]
print(hrefs)

It prints the following output :

['https://www.indigo.com/registration.html']
[]
['https://www.indigo.com/buservfcl.html', 'https://www.indigo.com/2021/07/agents.html']

getDetails has the complete page source code

Now, how do I compare getDetails with the hrefs list and remove/decompose every items that is present in the list.

I tried this , but it doesnt work for some reason :

hrefs = [links['href'] for links in getDetails.find_all('a', href=True) if target in links['href']]
print(hrefs)
for z in hrefs:
    getDetails.decompose()

It removed the entire data in the getDescription, but i need to remove only the elements which are in the list and not evrything

The output should be the complete HTML except the ones that has www.indigo.com in it

Answer 1

You have to find parent tag and then use decompose() method

html="""<div><a href="www.indigo.com"></div>"""

soup=BeautifulSoup(html,"html.parser")

target= "www.indigo.com"
href_tags = [links for links in soup.find_all('a', href=True) if target in links['href']]

for i in href_tags:
    i.parent.decompose()

Output:

soup will be empty

From the URL:

import requests
res=requests.get("https://www.assamcareer.com/2021/06/oil-india-limited.html")
soup=BeautifulSoup(res.text,"html.parser")
target= "www.assamcareer.com"
tags = [links for links in soup.find_all('a', href=True) if target in links['href']]
for i in tags:
    i.parent.decompose()

Updated Answer:

for title in root:
    /
 
        Your code

    /
    href_tags = [links for links in getDetails.find_all('a',href=True) if target in links['href']]
    print(href_tags)

for i in href_tags:
    i.parent.decompose()