MySQL - display image?

[mcfmullen]

I've seen a lot of information regarding storing a complete URL into MySQL an then calling it through php to show an image. This makes sense to me.

 

What I want to know, and I'm sure it will be easy to answer, is if I can only store a PART of a url in MySQL and call it through php...

 

EX: Image url is http://www.website.com/directory/image.png

 

Instead of storing that into a field in MySQL, I want to store ONLY image.png I then want to use php to call that bit into a given URL.

 

The reason being that if ever I relocate the image directory or change domain, I can simply edit the php file once rather than rename all of the values in MySQL.

 

So the question is: how can I call the variable into an incomplete URL in order to make the image display?

[taquitosensei]

$image="image.png"  // this is the image stored in your database 
$path="/var/www/html/images/";   this is the path to your images folder 

<img src='<?php echo $path.$image; ?>'>

[teamatomic]

A url would display an image, a path is broken.

 

$img='img.png';

$url='http://domain.com/';

echo "<img src=\"$url$img\">

 

 

HTH

Teamatomic

[mcfmullen]

The image.png part is actually stored in mysql:

 

I have this:

$url='http://www.mydomain.com/';

 

echo "Photo: <img src=\"$url".$row['photoAnimal'].\"<br/>";

 

But I'm getting errors. Can anyone correct my syntax? Thank you.

[mapleleaf]

Try this:

echo 'Photo: <img src="'.$url.$row['photoAnimal'].'"><br/>';

[mcfmullen]

I still get a syntax error. The code is part of a larger string being called by a MySQL table through an array with a loop:

 

$result = mysql_query("SELECT * FROM Animals");

$url='http://www.mydomain.com/';

 

while($row = mysql_fetch_array($result)){

echo "ID: ".$row['idAnimal'].", Photo: <img src="'.$url.$row['photoAnimal'].'">, Name: ".$row['nameAnimal']."<br/>";

}

 

No matter where I put the ', I keep getting a syntax error. Any advice?

[sader]

echo "ID: ".$row['idAnimal'].", Photo: <img src='".$url.$row['photoAnimal']."'>, Name: ".$row['nameAnimal']."<br/>";

[mcfmullen]

Success, thank you!

[mcfmullen]

I've been trying to get my information to show in a table:

 

$result = mysql_query("SELECT * FROM Animals");
$url='http://www.mcfilmmakers.com/wp-content/farmville/animals/';

$results_array = array(); 
while ($row = mysql_fetch_array($result)) { $results_array[$row['idAnimal']] = $row; }

?>

<table> 
<?php foreach ($results_array as $id => $record) { ?> 
<tr>
<td>
<?php echo $id;?>
</td>
<td>
[color=red]<img src='<?php echo $url$row['photoAnimal']; ?>'>[/color]
</td>
<td>
<?php echo $record['nameAnimal'];?>
</td>
</tr> 
<?php } ?> 
</table>

 

The result won't display what is inside the $row['photoAnimal'] variable thus giving me broken image links. $id and $record['nameAnimal'] display just fine. Any pointers? I've tried using the other suggestions, even the one that had worked in my previous attempt, but even that won't work here.

[mapleleaf]

<?php echo $url$row['photoAnimal']; ?>

should be:

<?php echo $url.$row['photoAnimal']; ?>