Dispaly Image With Php & Mysql

[jalaladdin]

i wrote this code for dispaly image but IE don't show it

<?php// connect to mysql server$link = mysql_connect('localhost', 'root', '');if (!$link) { die('Not connected : ' . mysql_error());}// connect to database server$db_selected = mysql_select_db('admin', $link); if (!$db_selected) { die ('Database error : ' . mysql_error());} $sql = mysql_query("SELECT 'path' FROM image WHERE id='3' "); while($row= mysql_fetch_array($sql)){ $image = $row['path']; // $image = $row['path']; echo'<img src="$image"/>';}?>

[jalaladdin]

please help me

[Don E]

For 'path', is the whole path to the image stored there or is it just the image name? In other words, does $image turn out to equal "me.jpg" or "images/me.jpg"? If it's just "me.jpg" then you have to specify the path to the image for the img src tag: echo '<img src="images/' . $image . ' " />'; Also, from my understanding.. when you query something from the database and it's a number(integer), like id='3', you don't need the single quotes. You can just do id=3.

[boen_robot]

$sql = mysql_query("SELECT 'path' FROM image WHERE id='3' ");

You should be getting the string "path", and your resulting code should be

<img src="path"/>

This would work if you have an image named "path" in the same folder as the PHP file, but that's not what you want I assume. To get the value of a column named "path", you need to use ` (backticks) instead of ' (apostrophes), or omit any `. In addition to that, when using numbers, you must omit the apostrophes. Assuming the "id" column is an integer, a more appropriate query would look like:

$sql = mysql_query("SELECT path FROM image WHERE id=3");