[Resolved] Just cannot display image..

kcgame

Hi all,

I've read thr all the 'display image from database' thread but still cannot solve my problem🙁

I've managed to store my image into db but now i have problems displaying it on browser. Below is a simplified version of my code, please take a look:


<?php
include("C:/topsecret.php");
		Header("Content-type: image/jpeg");

	// shows some row details here in table format. If i delete the following line, then the codes can display the image
	echo"<table border><tr><td>Hello</td></tr></table>";		

	// Connect to db
	$link=@mysql_pconnect($mysqlhost,$mysqluser,$mysqlpasswd);
	if($link==FALSE){
             echo "<p><b>Unfortuantely, no link to the database can be made. Therefore, no results. Try later.</body></html>\n";
             exit();
    	        }					
	mysql_select_db($mysqldbname,$link);

	//select respective image
	$realm_banner=mysql_query("SELECT * FROM image WHERE Rid = '8'" ,$link ) or die(mysql_error());		
	$pic=mysql_fetch_array($realm_banner);    	
	$data = $pic['Content']; 
   	$type = $pic['Type']; 
	echo "$data";

?>

The code can display the image perfectly if i delete away the table tag but this is impossible because i need to show the picture below the table. And i know that the Header have to be output before any information is passed to the browser, right?

Would appreciate any help/comment. Thanks.

paulnaj

HI kc,

Maybe this'll work ... create a file, image.php say ...

<?php 
// File:image.php

include("C:/topsecret.php"); 
if(!isset($_REQUEST['Rid'])){
	$_REQUEST['Rid'] = 8; // Or some other default
}
// Connect to db 
$link = mysql_pconnect($mysqlhost, $mysqluser, $mysqlpasswd)
	or die(mysql_error());

mysql_select_db($mysqldbname, $link)
	or die(mysql_error());

//select respective image 
$sql = "SELECT * FROM image WHERE Rid = '".$_REQUEST['Rid']."'";
$result = mysql_query($sql, $link)
	or die(mysql_error());    

if(!mysql_numrows($result)){
	die('No such image exists');
}
$pic = mysql_fetch_array($result);

header("Content-type: image/".$pic['Type']); 
echo $pic['Content']; 
?>

... which just delivers an image, according to a variable passed to it (in this case $Rid, but whatever).

Then, whenever you want an image from the database, you use ...

<img src="image.php?Rid=8">

... and you can put that wherever you need.

The only issue is what happens when something goes wrong (one of those 'die' statements occurs). I would consider returning a default image, maybe using fopen on a static jpg on the server and echo the contents instead of $pic['Content']. Plenty in the manual on how to do this.

Hope that helps.

Paul 🙂

kcgame

Thanks Paul😃 😃
Can't say enough to thank for your valuable help🙂
Now my picture is loading fine together with the rest of my webpage