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]]
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']
RegExp: \w+( \w+)+(?=]])
input
[[Alexander of Paris|poet named Alexander]]
output
poet named Alexander
input
[[Alexander of Paris]]
output
Alexander of Paris
import re
pattern = re.compile(r"\[\[([\w ]+)(?:\||\]\])")
text = "of which [[Alexander the Great]] was somewhat like [[King Arthur|Arthur]]"
results = pattern.findall(text)
print results
Would give the output
["Alexander the Great", "King Arthur"]
If you are trying to get all the links from a page, of course it is much easier to use the MediaWiki API if at all possible, e.g. http://en.wikipedia.org/w/api.php?action=query&prop=links&titles=Stack_Overflow_(website).
Note that both these methods miss links embedded in templates.
