Returning a Image file

[Daveyz1983]

how do i return a image file when someone access the page with a query string?

for example i se ebefore
<img src="something.php?img=118">

How does something.php return the image?

can someone guide me.

[stuartc1]

this is often used to retreive an image from a database (stored as a blob) - here is an example:

// querystring = page.php?id=21

mysql_connect("localhost","root","password");
mysql_select_db("prod");
// retrive binary image, and type
$query = "select bin_type, bin_data from $table where ref = '$id'";
$result = @MYSQL_QUERY($query);
$data = @MYSQL_RESULT($result,0,"bin_data");
$type = @MYSQL_RESULT($result,0,"bin_type");
$type = "";
// pass the image type and binary data to browser
Header("Content-type: $type");
// header("Content-Disposition: attachment; filename=$filename");
echo $data;

You need to have the binary data (image) and its type (application/jpeg or whatever) and pass these as headers to the browser.
In may be possible to do the same thing with an image file directly, you could give it a try.

[sleepingdanny]

This Will Work :p

File: "something.php"

You Just Need To Put A Link To:
<a href="something.php?img=118">TEXT</a>
And It Will Work.

<?
if($img=="118"){
print "<img src=http://www.site.com/118.gif>";
}
elseif($img=="245"){
print "<img src=http://www.site.com/245.gif>";
?>

[Inkybro]

You could also do:

echo "<img src=\"".$img."\">";

[cockney]

it's also used when a graphic is generated by a php script using GD

such as

<?php

$im = ImageCreateFromJpeg("images/banner.jpg");
// do something with image

Header("Content-Type: image/png");
ImagePNG($im);
ImageDestroy($im);
?>

[Daveyz1983]

got it, thanx for the replies guys.