<? $qu1=mysql_query("select * from news by id desc limit $num1"); while($a1=mysql_fetch_array($qu1)){ //echo "$al[title]"; $len1=strlen($a1["title"]); if($len1<="16") $leng1=$a1["title"]; else $leng1=substr($a1["title"],"0","60")."..."; echo "$leng1"; } ?> 为何出现MYSQL提供的变量无效！ 错误信息： Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /wwwroot/site/phphome/show.php on line 13