我在显示包含回复的推文时遇到了一个问题（例如，当你@username某人时）。它显示了@符号，但随后切断了推文的其余部分。（签出www.teamworksdesign.com 在右下角）

有谁能帮上忙或者推荐一个好的插件吗？

<?php

function parseTweet($text) {
$pattern_url = \'~(?>[a-z+]{2,}://|www\\.)(?:[a-z0-9]+(?:\\.[a-z0-9]+)?@)?(?:(?:[a-z](?:[a-z0-9]|(?<!-)-)*[a-z0-9])(?:\\.[a-z](?:[a-z0-9]|(?<!-)-)*[a-z0-9])+|(?:(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\\.){3}(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?))(?:/[^\\\\/:?*"|\\n]*[a-z0-9])*/?(?:\\?[a-z0-9_.%]+(?:=[a-z0-9_.%:/+-]*)?(?:&[a-z0-9_.%]+(?:=[a-z0-9_.%:/+-]*)?)*)?(?:#[a-z0-9_%.]+)?~i\';
\'@([A-Za-z0-9_]+)\';
$tweet = preg_replace(\'/(^|\\s)#(\\w+)/\', \'\\1#<a
href="http://search.twitter.com/search?q=%23\\2″ rel="nofollow">\\2</a>\', $text);
$tweet = preg_replace(\'/(^|\\s)@(\\w+)/\', \'\\1@<a
href="http://www.twitter.com/\\2″ rel="nofollow">\\2</a>\', $tweet);
$tweet = preg_replace("#(^|[\\n ])(([\\w]+?://[\\w\\#$%&~.\\-;:=,?@\\[\\]+]*)(/[\\w\\#$%&~/.\\-;:=,?@\\[\\]+]*)?)#is", "\\\\1<a
href=\\"\\\\2\\" title=\\"\\\\2\\" rel=\\"nofollow\\">[link]</a>", $tweet);
return $tweet;
}

$username=\'teamworksdesign\'; // set user name
$format=\'json\'; // set format
$tweet=json_decode(file_get_contents("http://api.twitter.com/1/statuses/user_timeline/{$username}.{$format}")); // get tweets and decode them into a variable

$theTweet = parseTweet($tweet[0]->text);
$newTweet = substr($theTweet,0,65);
echo \'<a class="tweet" rel="nofollow" href="http://www.twitter.com/teamworksdesign"> "\' . $newTweet . \'..."</a>\';

?>