正则表达式，略过几句话

Question

I am struggling to just get the text in between the quotes in the 'alt' tag. I have been trying regular expressions like [!?border="0"] to skip over it but still won't work.

I have tried \\s(border="0")\\s(alt=").*?" but it highlights over the 'border' tag

Here's the text that i'm trying to extract from using regex

<img src="http://www.ebgames.com.au/0141/169/5.png"alt="Far Cry 3" title=" Far Cry 3 " class="photo"/>            </a>

I am just trying to extract the text in between the quotes of the alt tag. Extracting the title would probably be better if possible. Please help, thank you

Answer 1

Try this regex:

border=\"0\" alt=\"(.*?)\"

Demo: https://regex101.com/r/1kbiBv/1/

You could also implement Positive Look-ahead, and Positive Look-behind to catch only what is between quotes:

(?<=border=\"0\" alt=\").*?(?=\")

Demo: https://regex101.com/r/1kbiBv/2/

Answer 2

There is better way to extract html element and attribute with BeautifulSoup :

from bs4 import BeautifulSoup
div_test='<img src="http://rcdn-1.fishpond.com.au/0141/169/297/319967448/5.jpeg" border="0" alt="The Durrells: Series 2" title=" The Durrells: Series 2 " class="photo"/> '
soup = BeautifulSoup(div_test, "lxml")
result = soup.find("img").get('alt')
result

Output:

'The Durrells: Series 2'

Answer 3

You can use a lambda in order to extract your tags from your current input.

You can try this code:

import re

a = '''<img src="http://rcdn-1.fishpond.com.au/0141/169/297/319967448/5.jpeg" border="0" alt="The Durrells: Series 2" title=" The Durrells: Series 2 " class="photo"/>            </a>
'''

find_tag = lambda x: r'{0}="(.*?)"'.format(x)
# Same as doing:
# regex = re.compile(find_tag('border="0" alt'))
regex = re.compile(find_tag("alt"))
text = re.findall(regex, a)
print(text)

Output:

['The Durrells: Series 2']

Also, this code will work with the other tags as well, for example:

regex = re.compile(find_tag("src"))
# Same as doing:
# regex = re.compile(find_tag('<img src'))
text = re.findall(regex, a)
print(text)

Output:

['http://rcdn-1.fishpond.com.au/0141/169/297/319967448/5.jpeg']

Answer 4

I think re.search with a simple regex works.

import re
s = '<img src="himg src="http://www.ebgames.com.au/0141/169/5.png" border="0" alt="Far Cry 3" title=" Far Cry 3 " class="photo"/>            </a>'
pat = 'alt="([^"]*)".* title="([^"]*)".*"'
a = re.search(pat, s)
print(a[1]) # content in the alt tag : "Far Cry 3"
print(a[2]) # content in the alt title : "Far Cry 3"

Answer 5

This code finds what you need, using this pattern: 'alt=".*?"' .

 import re

 w ='<img src="http://rcdn-1.fishpond.com.au/0141/169/297/319967448/5.jpeg" border="0" alt="The 
 Durrells: Series 2" title=" The Durrells: Series 2 " class="photo"/>   </a>'

 pattern = 'alt=".*?"'
 m = re.findall(pattern, w)
 print(m)