Question

我在尝试使用从JavaScript标记引用的扩展PHP文件来获取我的网站标题上的推文时遇到了一些麻烦 - 我在PHP文件中包含了标头引用，并且输出作为最后一行是javascript包含回声。

<?php
Header("content-type: application/x-javascript");
//Get Latest Tweet
function latest_tweet($username,$tweetnumber){
   $url = "http://search.twitter.com/search.atom?q=from:$username&amp;rpp=10";
   $xml = simplexml_load_file($url);
   $tweettitle = $xml->entry[$tweetnumber]->title;
   $mytweet = $xml->entry[$tweetnumber]->content;
   $firstChar = substr($tweettitle, 0, 1);
   //Exclude @ replies
   if($firstChar == "@"){
      //If this tweet is an @ reply move on to the previous one
      while ($firstChar == "@"){
         $tweetnumber++;
         $tweettitle = $xml->entry[$tweetnumber]->title;
         $mytweet = $xml->entry[$tweetnumber]->content;
         $firstChar = substr($tweettitle, 0, 1);
         if($firstChar != "@"){
            //If the previous tweet is not an @ reply output it
            return $mytweet;
         }
      }
   } else {
      //If first tweet is not an @ reply output it
      return $mytweet;
   }
}
//End Get Latest Tweet

//output
echo "document.write(latest_tweet('mikedeveloper', 0))";
?>

Answer 1

由于latest_tweet是一个PHP函数，你必须从PHP调用它，但是你试图从javascript调用它。试试这个：

echo "document.write('" . latest_tweet('mikedeveloper', 0) . "')";

从ext JavaScript文件引用PHP

1 个答案: