幫忙看看一下資料庫連接的錯誤！！！
[email protected]_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 $image;
?>
和
if($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 $i == 5)
{echo "";}$i++;
}
}
?>
------解決方案--------------------
都是由於mysql_db_query的執行結果出錯的原因
檢測mysql_connect("localhost","root","Admin","test"); 是不是有錯，[email protected]
還有就是執行的sql出錯的可能性比較大。
------解決方案--------------------
$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);
這樣看看。
