img and php

hey guys, how can i show images by php?

i mean

image;

<img src="http://www.site.com/img/324.jpg" />

Code (markup):

i want to use it like;

<img src="http://www.site.com/image.php?id=324" />

Code (markup):

what's the image.php?

<?php
$id = intval($_GET['id']);
header('Content-Type: image/jpeg');

// i couldn't do after this line

"http://www.site.com/img/".$id.".jpg";    // ??

?>

Code (markup):

 

Try this:

$id = intval($_GET['id']);
$the_image_file = "/img/".$id.".jpg";
$image = imagecreatefromjpeg($the_image_file);
header('Content-type: image/jpeg');
imagejpeg($image);
imagedestroy($image);

Code (markup):

 

No Need man its simple...!

<?php
echo "<img src='your image.jpg' width='300' height='500'>"
?>

 

Any html can be displayed in the browser by using echo or print. Like
<?php
$id = intval($_GET['id']);
echo "<img src='http://www.site.com/img/".$id.".jpg'>"
?>

or

<?php
$id = intval($_GET['id']);
print "<img src='http://www.site.com/img/".$id.".jpg'>"
?>

 

Though I am not sure but I think you're lacking a closing quotation mark and a semi-colon on this line:

 

On the last line before php closing tags, its not needed to put semi-colon after the code. Refer http://php.net/manual/en/language.basic-syntax.instruction-separation.php

 

Oops! The Guy is talking about showing an image using a Dynamic PHP Page template

BTW: The Thread Starter imgae.php must look like

<?php
$id = intval($_GET['id']);
header('Content-Type: image/jpeg');

// i couldn't do after this line
?>
<html><head></head><body>
<img src="http://www.site.com/img/<?php echo $id; ?>.jpg" />
<body></html>
?>

PHP:

 

thanks it works very well