Question

How do I do a replace that will turn all plain addresses in a paragraph in to links?

The problem is periods are valid in the address, but the address may be at the end of a sentence.

Test string:

The link is: http://www.google.com/pants.  And that is the link.  

I need to group out from http to pants. (It is plain text and I need to make it html.)

This regex grabs the period after pants and so is bad:

(^|[\n ])([\w]+?://[^\s]*)

I’m pretty sure I need to do a lookahead, but I can’t put that in the ^\s character set. Trying to do an ifthenelse has also eluded me.

Here is my output thingy:

$1<a href=\"$2\" target=\"_blank\">$2</a>

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Answer 1

Assuming no space in the urls and a space or end of string after them:

str = str.replace(
    /(https?:\/\/\S+?)(?=\.?(\s|$))/g,
        '<a href="$1" target="_blank">$1</a>' );

It captures ‘http[s]://’ and non-space characters as few times as possible until looking ahead there is optionally . and then a space or the end of string.

If you want to exclude other punctuation that may be at the end of an url you could change the positve lookahead accordingly, e.g. (?=[;:!,.]?(?:\s|$)).

Note that the above regex is not intended to only match valid urls, and you may want to replace the \S with [\w/.-] to only match urls containing word characters and .-/.

In search of the perfect URL validation regex