regular expression

jbjaz posted this at 21:29 — 27th May 2005.

They have: 4 posts

Joined: May 2005

Hi, I hope someone can help me.

I'm using php.

Basically, I've captured a bunch of html in a variable, we'll call it $html

Now, I want to parse through the html and find every instance of
[some text]
then I want to look through "some text" and find all newlines (\n) and convert them to 's.

So, $html = preg_replace("reg exp","",$html);

I've been scratching my head and I just can't figure it out.

Thanks in advance.

Busy posted this at 22:18 — 27th May 2005.

He has: 6,151 posts

Joined: May 2001

<?php
$newhtml = eregi_replace(quotemeta(\"<address>\\0</address>\"),\"<address>\\0</address>\",$html);
/* doesn't really do anything, just converts the section to variable, will need 
an array or loop if more than one instance */
$newhtml = str_replace(\"\n\",\"<br />\",$newhtml);
/* $newhtml should be the orginal value with changes made */
?>

Try that, I haven't tested it but should work, note quotemeta() and \\0 should remain
Could also be done in a loop, but seems you only want to change \n to in certain areas

jbjaaz posted this at 13:40 — 31st May 2005.

They have: 6 posts

Joined: Apr 2005

It almost works. It adds s to every newline regardless of where they are. So

hello world
<address>
23 blah
blah blah
</address>

'
becomes

hello world<br>
<address><br>
23 blah<br>
blah blah<br>
</address><br>

Thanks tho. Someone gave me a solution on another forum...

$string = <<<END
<address>
You and Yours
Your Home Address
Your Home State
</address>
END;

function ins_BR($arr){
return $arr[1].str_replace("\r\n", "<br />\n",$arr[2]).$arr[3];
}

$pattern = "/(<address>\r\n)+([^<]+)+(<\/address>)/";
echo preg_replace_callback($pattern, "ins_BR", $string);

'
which works fine excepts it adds a to the last line between the tags. If there are two lines, only the first should get the .

Mark Hensler posted this at 04:25 — 16th June 2005.

He has: 4,048 posts

Joined: Aug 2000

try:

<?php
echo preg_replace_callback(\"!<address>\s*(.+)\s*</address>!isU\", \"nl2br\", $string);
?>