I'm just starting regular expression for python and came across this problem where I'm supposed to extract URLs from the string:
str = "<tag>http://example-1.com</tag><tag>http://example-2.com</tag>"
The code I have is:
import re
url = re.findall('<tag>(.*)</tag>', str)
print(url)
returns:
[http://example-1.com</tag><tag>http://example-2.com]
If anyone could point me in the direction on how I might approach this problem would it would be most appreciative!
Thanks everyone!
You are using a regular expression, and matching HTML with such expressions get too complicated, too fast.
You can use BeautifulSoup to parse HTML.
For example:
from bs4 import BeautifulSoup
str = "<tag>http://example-1.com</tag><tag>http://example-2.com</tag>"
soup = BeautifulSoup(str, 'html.parser')
tags = soup.find_all('tag')
for tag in tags:
print tag.text
Using only re package:
import re
str = "<tag>http://example-1.com</tag><tag>http://example-2.com</tag>"
url = re.findall('<tag>(.*?)</tag>', str)
print(url)
returns:
['http://example-1.com', 'http://example-2.com']
Hope it helps!
.*? is matches as few times as possible (lazy), i.e. when it finds the first closing </tag> it stops, whereas in the greedy .* as soon it finds the 2nd closing </tag> it matches the whole pattern.
.*?non-greedy instead of.*greedy one Or use[^>]*instead of.*OR best use a HTML parser