CODEWITHSUNDEEP
pythonhtmlmarkdown
Erica Lan
Erica Lan

Reputation: 57

Convert hyperlink

I am trying to write a python function and convert markdown format link like: [text] (link), into < a href > tags in HTML. For example:

link(line)

link("Here is the link [Social Science Illustrated](https://en.wikipedia.org/wiki/Social_Science_Illustrated) I gave you yesterday.")

"Here is the link <a href="https://en.wikipedia.org/wiki/Social_Science_Illustrated">Social Science Illustrated</a> I gave you yesterday."

And my code for now is:

def link(line):
  import re
  urls = re.compile(r"((https?):((//)|(\\\\))+[\w\d:#@%/;$()~_?\+-=\\\.&]*)")
  line = urls.sub(r'<a href="\1"></a>', line)
  return line

Output:

=> 'Here is the link [Social Science Illustrated] ( <a href="https://en.wikipedia.org/wiki/Social_Science_Illustrated"></a> ) I gave you yesterday.'

So I was wondering how to convert [text] part into correct position?

Upvotes: 1

Views: 1871

Answers (2)

Anurag Misra
Anurag Misra

Reputation: 1544

from re import compile, sub

def html_archor_tag(match_obj):
    return '<a href="%(link)s">%(text)s</a>' %{'text': match_obj.group(1), 'link': match_obj.group(2)}

def link(line):
    markdown_url_re = re.compile(r'\[([^\]]*)]\(([^\)]*)\)')
    result = sub(markdown_url_re, html_archor_tag, line)
    return line

For more information on re.sub

Upvotes: 0

aaron
aaron

Reputation: 43073

If you just need to convert based on the [text](link) syntax:

def link(line):
  import re
  urls = re.compile(r'\[([^\]]*)]\(([^\)]*)\)')
  line = urls.sub(r'<a href="\2">\1</a>', line)
  return line

You don't have to validate the link. Any decent browser will not render it as a link.

Upvotes: 1

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