php regex - Replace all @usernames with link in a string

Question

I am implimenting a timeline system on my website where users can mention others users in their timeline using @username like twitter.

I want to convert @username to link and point it to their profile

My string :

$timeline="@fred-ii 's posts on @stackoverflow are intresting.";

I am using the following code to replace @username with url :

echo preg_replace("/@([^\s]+)/i","<a href='http://example.com/$1'>@$1</a>",$timeline);

it works, the problem is that it matchs spaces also

this string

"@fred-ii 's posts on@stackoverflow";

There is no space between on and @stackoverflow ,I want to exclude it,

so i updated my regex

/\s+@([^\s]+)/

it worked but it didnot match the first part of my string (username @fred-ii ) .I think regex engine is looking for space|s at the start of string.

What do I need to change in my pattern to match all @usernames ?

$timeline="@fred-ii 's posts on @stackoverflow are intresting.";

So you want to match the space in the username, but don't have a space in the link? — Shomz
– Shomz, Commented Nov 21, 2015 at 15:05

Marcos Dimitrio · Accepted Answer · 2015-11-21 15:20:51Z

2

You can do this with lookbehind negative assertion:

/(?<!\w)@([^\s]+)/

The (?<!\w) tells the Regex engine to match @([^\s]+) only if it's not preceded by a word \w. It will work in the example you gave, perhaps you will have to tweak it as you go.

Example code:

$pattern = "/(?<!\w)@([^\s]+)/";
$subject = "@fred-ii 's posts on @stackoverflow are interesting. @fred-ii 's posts on@stackoverflow";

preg_match_all($pattern, $subject, $matches, PREG_SET_ORDER );

foreach($matches as $item)
{
    echo $item[1] . "<br/>";
}

Produces this output:

fred-ii
stackoverflow
fred-ii

See it in action.

answered Nov 21, 2015 at 15:20

Marcos Dimitrio

6,9306 gold badges42 silver badges64 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

anubhava · Accepted Answer · 2015-11-22 08:21:56Z

0

You can use this lookbehind assertion:

/(?<=\s|^)@\S+/

(?<=\s|^) makes sure there is whitespace or line start before @.

Code:

php> $s = "@fred-ii 's posts on@stackoverflow";

php> echo preg_replace('/(?<=\s|^)@\S+/', "<a href='http://example.com/$0'>$0</a>", $s);
<a href='http://example.com/fred-ii'>@fred-ii</a> 's posts on@stackoverflow

edited Nov 22, 2015 at 8:21

answered Nov 21, 2015 at 15:24

anubhava

790k67 gold badges603 silver badges671 bronze badges

5 Comments

Amit Verma Over a year ago

Anubhava your answer is korrect, it gives the korrect output. but what if I don't use lookbehind forexample the pattern /(\s+|^)@([^\s]+)/ also works, is it correct to use it or always with lookbehind. please guide me.

anubhava Over a year ago

Yes you can also use: /(?:\s|^)@([^\s]+)/ with captured group. Lookbehind gives you an option of not using any captured group at all using: preg_replace('/(?<=\s|^)@\S+/', "<a href='http://example.com/$0'>$0</a>", $s);

Amit Verma Over a year ago

Thank you so much for the explaination Anubhava , I learned alot about LookBehind from your answer and comment. I really appriciate it.

Amit Verma Over a year ago

Anubhava Do you know how to match a string in Shell based wildcard expresion? sorry for adding an unrelated comment, I'ma remove it soon.

anubhava Over a year ago

Try: s="@fred-ii 's posts on@stackoverflow"; sed -r 's~(^|[[:blank:]])(@[^[:blank:]]+)~\1<a href="http://example.com/\2">\2</a>~' <<< "$s"

Collectives™ on Stack Overflow

php regex - Replace all @usernames with link in a string

2 Answers 2

Comments

5 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

5 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related