帮忙看看一下数据库连接的错误！！！
<?php
$connid=@mysql_connect("localhost","root","Admin","test");
$sql="select pic from $table where id=$id";
$result=mysql_db_query("photo",$sql);
$image=mysql_result($result,0,"pic");
echo $image;
?>
和
<?php
if($offset == "")
{$offset = 0;}
$Connid = @mysql_connect("localhost","root","Admin","test");
$sql = "select id,name from photo order by id desc limit $offset,6";
$result = mysql_db_query("photo",$sql);
$num = mysql_num_rows($result);
if($num != 0)
{
$i = 0;
while($r = mysql_fetch_array($result))
{
$id = $r["id"];
$ftitle = $r["name"];
if($i == 0 or $i == 3)
{
echo "
}
echo "
echo "";
echo "show.php?table=photo&id=$fid";
echo "height=60>";
echo "";
echo "
$ftitle
";if($i == 2 or $i == 5)
{echo "
}
}
?>
------解决方案--------------------
都是由于mysql_db_query的执行结果出错的原因
检测mysql_connect("localhost","root","Admin","test"); 是不是有错，把@去掉。
还有就是执行的sql出错的可能性比较大。
------解决方案--------------------
$Connid = mysql_connect("localhost","root","Admin");
mysql_select_db（"photo"）；
$sql = "select id,name from photo order by id desc limit $offset,6";
$result = mysql_query($sql);
$num = mysql_num_rows($result);
这样看看。