Question

If I have some xml containing things like the following mediawiki markup:

” …collected in the 12th century, of which [[Alexander the Great]] was the
hero, and in which he was represented,
somewhat like the British [[King
Arthur|Arthur]]”

what would be the appropriate arguments to something like:

re.findall([[__?__]], article_entry)

I am stumbling a bit on escaping the double square brackets, and getting the proper link for text like: [[Alexander of Paris|poet named Alexander]]

Answer 1

Here is an example

import re

pattern = re.compile(r"\[\[([\w \|]+)\]\]")
text = "blah blah [[Alexander of Paris|poet named Alexander]] bldfkas"
results = pattern.findall(text)

output = []
for link in results:
    output.append(link.split("|")[0])

# outputs ['Alexander of Paris']

Version 2, puts more into the regex, but as a result, changes the output:

import re

pattern = re.compile(r"\[\[([\w ]+)(\|[\w ]+)?\]\]")
text = "[[a|b]] fdkjf [[c|d]] fjdsj [[efg]]"
results = pattern.findall(text)

# outputs [('a', '|b'), ('c', '|d'), ('efg', '')]

print [link[0] for link in results]

# outputs ['a', 'c', 'efg']

Version 3, if you only want the link without the title.

pattern = re.compile(r"\[\[([\w ]+)(?:\|[\w ]+)?\]\]")
text = "[[a|b]] fdkjf [[c|d]] fjdsj [[efg]]"
results = pattern.findall(text)

# outputs ['a', 'c', 'efg']