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#2
风月_无边2008-11-10 16:32
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<?
require_once("connect_sql.php");
$sql="select * from table_2 ";
$result=mysql_query($sql,$link) or die("failed to display the picture");
while($row=mysql_fetch_array($result))
{
$data=base64_decode($row['logo']);
header("Content-type: image/jpeg");
echo $data;
}
//$data=base64_decode($row['logo']);
//header("Content-type: image/jpeg");
//echo $data;
?>
require_once("connect_sql.php");
$sql="select * from table_2 ";
$result=mysql_query($sql,$link) or die("failed to display the picture");
while($row=mysql_fetch_array($result))
{
$data=base64_decode($row['logo']);
header("Content-type: image/jpeg");
echo $data;
}
//$data=base64_decode($row['logo']);
//header("Content-type: image/jpeg");
//echo $data;
?>
我的代码只能显示一张图片
但是我想将我的数据库中的图片都显示在一个页面上,总是出现警告和乱码!!不知道是为什么!!请高人指点!
总是显示刚插入数据库的图片!!