regular expression in string.replace with a callback function

Question

 function helpLinkConvert(str, p1, offset, s)  {  
      return "<a href=\"look.php?word="
               +encodeURIComponent(p1)+"\">"+p1+"</a>";
     }

var message = "(look: this) is a (look: stackoverflow) question";
message = message .replace(/\(look: (.{1,80})\)/, helpLinkConvert);

This is what I want to do,

Before:

(look: this) is a (look: stackoverflow) question.

After:

this is a stackoverflow question

When there is only one matched string, it's working but in other cases It's not working properly,

How can I do that? Thanks.

Paul · Accepted Answer · 2011-12-10 20:03:07Z

5

You need to add the global g modifier , and a non-greedy match so the regular expression finds all matches:

/\(look: (.{1,80}?)\)/g

In your code:

function helpLinkConvert(str, p1, offset, s) {  
    return "<a href=\"look.php?word="+encodeURIComponent(p1)+"\">"+p1+"</a>";
}

var message = "(look: this) is a (look: stackoverflow) question";
message = message.replace(/\(look: (.{1,80}?)\)/g, helpLinkConvert);

Outputs:

"<a href="look.php?word=this">this</a> is a <a href="look.php?word=stackoverflow">stackoverflow</a> question"

edited Dec 10, 2011 at 20:03

answered Dec 10, 2011 at 19:34

Paul

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1 Comment

ridgerunner Over a year ago

Or better yet, instead of: .{1,80}? use: [^)]{1,80}

Vivin Paliath · Accepted Answer · 2011-12-10 19:35:37Z

2

Use the g flag:

message .replace(/\(look: (.{1,80})\)/g, helpLinkConvert);

The g (stands for "global") will match against all occurrences of the pattern on this string, instead of just the first one.

answered Dec 10, 2011 at 19:35

Vivin Paliath

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