如何使用PHP显示存储在mysql数据库中的图像
I have been trying to display image from database using php but i am only getting download prompt whenever i use or click the link. what i want to do is, i want to display it on the web browser. And want to use it in tags.
my code is:
require_once('dbconfig.php');
$cont=mysql_connect($dbhost,$dbuser,$dbpass) or die("Error Connecting the Database");
$seldatabase=mysql_select_db($dbselect,$cont);
$insert = "SELECT `p_mime`, `p_name`, `p_size`, `p_data` FROM $dbtbl_reg_details WHERE `id` = $id";
$query = mysql_query($insert);
if ($query)
{
if (mysql_num_rows($query)==1)
{
$row = mysql_fetch_assoc($query);
header("Content-Type: image/png");
header("Content-Length: ". $row['p_size']);
header("Content-Disposition: inline; filename=". $row['p_name']);
echo $row[p_data];
}
else
{
echo "image with id = ".$id." does not exist";
}
}
else
{
echo "query failed, image with id = ".$id." does not exist";
}
And when i use inline for Content-Disposition: then my web page returns a script error on the browser.
So, what should i do to display images on web pages while retrieving it from mysql database
我一直在尝试使用php从数据库显示图像,但我只是在使用或点击 链接。 我想要做的是,我想在网络浏览器上显示它。 并希望在标签中使用它。 p>
我的代码是: p>
require_once('dbconfig.php');
$ cont = mysql_connect($ dbhost,$ dbuser,$ dbpass)或die(“连接数据库时出错”);
$ seldatabase = mysql_select_db($ dbselect,$ cont);
$ insert =“SELECT`p_mime`,`p_name `,`p_size`,`p_data` FROM $ dbtbl_reg_details WHERE`id` = $ id“;
$ query = mysql_query($ insert);
if($ query)
{
if if(mysql_num_rows($ query) )== 1)
{
$ row = mysql_fetch_assoc($ query);
header(“Content-Type:image / png”);
header(“Content-Length:”。$ row ['p_size ']);
header(“Content-Disposition:inline; filename =”。$ row ['p_name']);
echo $ row [p_data];
}
else
{
echo“ id =“。$ id。”的图像不存在“;
}
}
\ {
呼叫”查询失败,图片ID =“。$ id。”不存在“; \ n}
code> pre> \ n
当我使用内联进行内容处理:然后我的网页在浏览器上返回一个脚本错误。 p>
那么,我该怎么做才能在网上显示图像 从mysql数据库中检索它时的页面 p>
div>
if (mysql_num_rows($query)==1)
{
while( @ob_end_clean() );
header("Content-type: image/png");
header("Content-Length: ". $row['p_size']);
header("Content-Disposition: attachment; filename=\"{$row['p_name']}\"");
die($row[p_data]);
}
why u have echo hi;
?
its probably cause an error
and also add exit in the end (to sure you dont echo anything after)
echo $row[p_data];
exit;