参数传递与参数引用问题??
我有一个flash播放器,用object标签引入到一个php页面里面,传入参数到这个php页面,然后用$_GET[""]来获得传入参数,然后将传入的参数整理成一个字符串,作为param里面的一个值,但是无论如何都实现不了。我是第一次接触php这个东西,请高手们指教一下,谢谢
function createValue($filename,$showtop,$showinfo,$showvolume)
{
$val="mp3=$filename&showtop=$showtop&showinfo=$showinfo&showvolume=$showvolume";
return $val;
}
?>
在这里不知道为什么就是不行,好像createValue函数返回的值根本就不是一个字符串似的。
------解决方案--------------------
- PHP code
function createValue($filename,$showtop,$showinfo,$showvolume) { $val="mp3=$filename&showtop=$showtop&showinfo=$showinfo&showvolume=$showvolume"; echo $val;//看看是什么? } <br><font color="#e78608">------解决方案--------------------</font><br>浏览器打开后,用查看源文件看一看<br><param name="FlashVars" value="<?php" createvalue> /><br>这一行解析出来的value是什么。 <div class="clear"> </div>

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