views:

700

answers:

2

Hello,

I am trying to do some stuff with replacing String containing some URL to a browser compatible linked URL.

My initial String looks like this :

"hello, i'm some text with an url like http://www.the-url.com/ and I need to have an hypertext link !"

What I want to get is a String looking like :

"hello, i'm some text with an url like <a href="http://www.the-url.com/"&gt;http://www.the-url.com/&lt;/a&gt; and I need to have an hypertext link !"

I can catch URL with this code line :

String withUrlString = myString.replaceAll(".*://[^<>[:space:]]+[[:alnum:]/]", "<a href=\"null\">HereWasAnURL</a>");

Maybe the regexp expression needs some correction, but it's working fine, need to test in further time.

So the question is how to keep the expression catched by the regexp and just add a what's needed to create the link : catched string

Thanks in advance for your interest and responses !

+3  A: 

Try to use:

myString.replaceAll("(.*://[^<>[:space:]]+[[:alnum:]/])", "<a href=\"$1\">HereWasAnURL</a>");

I didn't check your regex.

By using () you can create groups. The $1 indicates the group index. $1 will replace the url.

I asked a simalir question: my question
Some exemples: Capturing Text in a Group in a regular expression

Martijn Courteaux
This works fine !Many thanks :D
Dough
+1  A: 

Assuming your regex works to capture the correct info, you can use backreferences in your substitution. See the Java regexp tutorial.

In that case, you'd do

myString.replaceAll(....., "<a href=\"\1\">\1</a>")
Matt