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Date: Wed, 25 Nov 2015 12:42:00 -0800 (PST)
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Subject: Screen scraper to get all 'a title' elements
From: ryguy7272 <ryanshuell@gmail.com>
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Xref: csiph.com comp.lang.python:99484

Hello experts.  I'm looking at this url:
https://en.wikipedia.org/wiki/Wikipedia:Unusual_place_names

I'm trying to figure out how to list all 'a title' elements.  For instance,=
 I see the following:
<a title=3D"Accident, Maryland" href=3D"/wiki/Accident,_Maryland">Accident<=
/a>
<a class=3D"new" title=3D"Ala-Lemu (page does not exist)" href=3D"/w/index.=
php?title=3DAla-Lemu&action=3Dedit&redlink=3D1">Ala-Lemu</a>
<a title=3D"Alert, Nunavut" href=3D"/wiki/Alert,_Nunavut">Alert</a>
<a title=3D"Apocalypse Peaks" href=3D"/wiki/Apocalypse_Peaks">Apocalypse Pe=
aks</a>

So, I tried putting a script together to get 'title'.  Here's my attempt.

import requests
import sys
from bs4 import BeautifulSoup

url =3D "https://en.wikipedia.org/wiki/Wikipedia:Unusual_place_names"    =
=20
source_code =3D requests.get(url)=20
plain_text =3D source_code.text
soup =3D BeautifulSoup(plain_text)
for link in soup.findAll('title'):
    print(link)

All that does is get the title of the page.  I tried to get the links from =
that url, with this script.

import urllib2
import re

#connect to a URL
website =3D urllib2.urlopen('https://en.wikipedia.org/wiki/Wikipedia:Unusua=
l_place_names')

#read html code
html =3D website.read()

#use re.findall to get all the links
links =3D re.findall('"((http|ftp)s?://.*?)"', html)

print links

That doesn't work wither.  Basically, I'd like to see this.

Accident
Ala-Lemu
Alert
Apocalypse Peaks
Athol
=C5
Barbecue
B=E5stad
Bastardstown
Batman
Bathmen (Battem), Netherlands
...
Worms
Yell
Zigzag
Zzyzx

How can I do that?
Thanks all!!