How to extract specific text in a list created by BeautifulSoup

Question

So I would like to get a list of download links from a page:

soup = BeautifulSoup(driver.page_source)
linky=soup.find_all(name='a', href=re.compile('download.php'))

This returns me a list of all the links:

[<a href="download.php/947983/adam.zip"><img "="" alt="Download" src="browse_dl.png" style="style=" title="Download Adam"/></a>,
<a href="download.php/947981/barb.zip"><img "="" alt="Download" src="browse_dl.png" style="style=" title="Download Barb"/></a>,
<a href="download.php/947972/chris.zip"><img "="" alt="Download" src="browse_dl.png" style="style=" title="Download Chris"/></a>,
<a href="download.php/947971/dan.zip"><img "="" alt="Download" src="browse_dl.png" style="style=" title="Download Dan"/></a>]

I would like to extract the href link, and the img title after the "Download", and then put them into tuples.

So I would have a list like the following:

[(download.php/947983/adam.zip, Adam)
(download.php/947981/barb.zip, Barb),
(download.php/947972/chris.zip, Chris),
(download.php/947971/dan.zip, Dan)]

I thought I could just split the text between href=" and "img for each item, but then I would have no idea how to do that, and the next problem is how would I also extract the title as well?

Answer 1

Here is a solution to your problem, let say we have a list_of_names containing your input links and names that you want to extract, then links and names can be extracted using code given below:

#!/usr/bin/python
import re
list_of_names= ['<a href="download.php/947983/adam.zip"><img "="" alt="Download"  src="browse_dl.png" style="style=" title="Download Adam"/></a>',
'<a href="download.php/947981/barb.zip"><img "="" alt="Download" src="browse_dl.png" style="style=" title="Download Barb"/></a>',
'<a href="download.php/947972/chris.zip"><img "="" alt="Download" src="browse_dl.png"  style="style=" title="Download Chris"/></a>',
     '<a href="download.php/947971/dan.zip"><img "="" alt="Download" src="browse_dl.png" style="style=" title="Download Dan"/></a>']

links=[]
names=[]


for row in list_of_names:
    links.append([x.strip() for x in re.split(r"href=\"(.*)\"><img", row)][1])
    names.append([x.strip() for x in re.split(r"title=\"Download (.*)\"\/>", row)][1])

desired_list=list(tuple(zip(links,names)))
print(desired_list)

If you compile this script then you can get your desired output:

python -i code_for_desired_output.py
[('download.php/947983/adam.zip', 'Adam'), ('download.php/947981/barb.zip', 'Barb'), ('download.php/947972/chris.zip', 'Chris'), ('download.php/947971/dan.zip', 'Dan')]