复制内容到剪贴板
代码:
<?php
//include("connect.php");
$db=mysql_connect("localhost","root","19871027");
mysql_select_db($guestbook,$db);
$result=mysql_query("SELECT*FROM guester",$db);
if ($myrow=mysql_fetch_array($result)){
echo'
\n';
do{
printf("%s,%s,%s",$myrow["name"],$myrow["email"],$myrow["blog"]);
echo"
\n";
printf("%s",$myrow["book"]);
}
while($myrow=mysql_fetch_arrar($result));
echo"
\n";
}
else{
echo"错误\n";
}
?>错误信息：
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in
C:\AppServ\www\guestbook\index.php on line
6
错误
各位大虾们，该如何修改呢？
