I am tring to extract a link from a phrase and it could be any where last, first or middle so I am usig this regex
link=text.scan(/(^| )(http.*)($| )/)
but the problem is when the link is in the middle it gets the whole phrase until the end. What should I do ?
It's because .* next to http is greedy. I suggest you to use lookarounds.
link=text.scan(/(?<!\S)(http\S+)(?!\S)/)
OR
link=text.scan(/(?<!\S)(http\S+)/)
Example:
> "http://bar.com foo http://bar.com bar http://bar.com".scan(/(?<!\S)http\S+(?!\S)/)
=> ["http://bar.com", "http://bar.com", "http://bar.com"]
(?<!\S) Negative lookbehind which asserts that the match won't be preceeded by a non-space character.
http\S+ Matches the substring http plus the following one or more non-space characters.
Do all the links you are trying to match follow some simple pattern? We'd need to see more context to confidently provide a good solution to your problem.
For example, the regex:
link=text.scan(/http.*\.com/)
...might be good enough for the job (this assumes all links end in ".com"), but I can't say for sure without more information.
Or again, for example, perhaps you could use something like:
link=text.scan(/http[a-z./:]*) - this assumes all links contain only lower case letters, ".", "/" and ":".
