Display image using php variable

Question

I am trying to display images by taking their path to the file from the sql table, but I have a lot of problems.

Here's what happens:

$imageis a variable containing text "itemimg/hyuna.png"that is the path to the image.

$image = 'itemimg/hyuna.png';

I assumed that I could display the image outside the php block as follows:

<img src= "<? $image ?>" alt="test"/>

For some reason this does not work.

So I thought that maybe it cannot read the variable outside of the php block (I'm a newbie), so for testing, I did:

<h1> "<? $image ?>" </h1>

It displays itemimg/hyuna.pngas an h1 banner.

This means that it is accessing the varible prefix.

So I thought the way was wrong. So I tried:

<img src= "itemimg/hyuna.png" alt="test"/>

This image perfectly displays the image.

, , , "test" "alt="

: sql ? :

$q = "select * from item where id=$id";
$results = mysql_query($q);
$row = mysql_fetch_array($results, MYSQL_ASSOC);
$image = ".$row['image'].";

item - : image,

Answer 1

, PHP Shorttags.

Shorttags PHP, :

<img src="<?=$image ?>" alt="test" />

:

<img src="<?php echo htmlspecialchars($image); ?>" alt="test" />

:

, , $image = $row['image'];

Answer 2

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

Answer 3

<img src= "<?= $image ?>" alt="test"/>

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

Answer 4

try it

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