Displaying an image using a php variable

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

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'];

Try this

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

try

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

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

try this

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

来源：https://stackoverflow.com/questions/19563639/displaying-an-image-using-a-php-variable

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