Displaying an image using a php variable - php

Question

I'm trying to display images by taking their file path from an sql table, but i'm having a lot of troubles.
Here is whats going on:
$image is a variable containing the text "itemimg/hyuna.png" which is path to an image.
$image = 'itemimg/hyuna.png';
I assumed I would be able to display the image outside of the php block like so:
<img src= "<? $image ?>" alt="test"/>
This doesn't work though for some reason.
So I thought maybe it's not able to read the variable outside the php block(i'm a beginner), so for testing i did:
<h1> "<? $image ?>" </h1>
It displays itemimg/hyuna.png as a h1 banner.
Meaning it's accessing the varible fine.
So I thought maybe the path is wrong. So I tried:
<img src= "itemimg/hyuna.png" alt="test"/>
This displays the image perfectly.
So now I'm stuck scratching my head why the first bit of code displays nothing but the text "test" from "alt="
Extra question:
How do I go about assigning a value from an sql cell to a variable?
I attempted the following with no luck:
$q = "select * from item where id=$id";
$results = mysql_query($q);
$row = mysql_fetch_array($results, MYSQL_ASSOC);
$image = ".$row['image'].";
item is a table with a collumn: image which contains file paths to images

Answer 1

First of all, you should not use PHP Shorttags.
When you use the PHP Shorttags you have to say:
<img src="<?=$image ?>" alt="test" />
But i would encourage to escape the Content off the variable like this:
<img src="<?php echo htmlspecialchars($image); ?>" alt="test" />
Your extra question:
This should lead to an syntax error because the string could not be parsed, just use $image = $row['image'];

Answer 2

Try this
<img src= "<?php echo $image ?>" alt="test"/>

Answer 3

try
<img src= "<?= $image ?>" alt="test"/>
or
<img src= "<? echo $image; ?>" alt="test"/>

Answer 4

try this
<img src= "<?php echo $image ?>" alt="test"/>