[SOLVED] Image not appeaaring

[mike12255]

I created some code to grab and image from X folder and by X name (both combined into one var called $image) i get no errors when i view the page, but the images don't appear. I tried echoing $name and it showed up properly as uploads/image.gif

 

heres my code :

 

<?php   
    
    $findimage = mysql_query ("Select * FROM `phpbb_files`");

while ($row = mysql_fetch_array($findimage)){
$image ="uploads/"; 
$image .= $row['shortname'];


?>
<img src= <?=$image?> />

<?php }?>

[DeanWhitehouse]

<?php   
    error_reporting(E_ALL);
    $findimage = mysql_query ("Select * FROM `phpbb_files`");
   
   while ($row = mysql_fetch_array($findimage)){
   $image ="./uploads/"; 
   $image .= $row['shortname'];

echo "<img src=\"".$image."\"  alt=\"Image can not be found\" />";
}
?>

Try that

[mike12255]

still not showing an image  :-\

[DeanWhitehouse]

It will say something then, what does it say?

[Sudden]

<img src= <?=$image?> />

 

should probly read

 

<img src= <?php $image ?> />

[DeanWhitehouse]

No it shouldn't

 

<?="anything"?>

The above requires short tags to be on

 

is like doing

<?php echo "anything";?>

[Prismatic]

change

 

$findimage = mysql_query ("Select * FROM `phpbb_files`");

 

to

$findimage = mysql_query ("Select * FROM `phpbb_files`") or die(mysql_error());

and see if anything is displayed

[DeanWhitehouse]

Going on that idea also add

echo mysql_num_rows($findimage);

under the query to see how many rows it finds

[mike12255]

I tried what you mates suggested so i've been editing it, and got the following error:

 

 

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/sgiclan/public_html/gnomes/images.php on line 466

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/sgiclan/public_html/gnomes/images.php on line 467

 

<?php   
    error_reporting(E_ALL);
   $findimage = mysql_query ("Select * FROM `phpbb_files`") or die(mysql_error());
   echo mysql_num_rows($findimage['shortname']);
   while ($row = mysql_fetch_array($findimage['shortname'])){
   $image ="uploads/"; 
   $image .= $row['shortname'];

echo "<img src=\"".$image."\"  alt=\"Image can not be found\" />";
}
?>

[DeanWhitehouse]

What the hell, is that your code?

 

Do this

<?php   
    error_reporting(E_ALL);

   $findimage = mysql_query ("Select * FROM `phpbb_files`") or die(mysql_error());
   echo mysql_num_rows($findimage);

   while ($row = mysql_fetch_array($findimage)){
   $image ="uploads/"; 
   $image .= $row['shortname'];

echo "<img src=\"".$image."\"  alt=\"Image can not be found\" />";
}
?>

And tell us what it says

[mike12255]

it said:

 

1
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/sgiclan/public_html/gnomes/images.php on line 469

[DeanWhitehouse]

I changed the code

[mike12255]

crap its back to saying nothing, its suposed to appear under the video player: http://sgi-clan.info/gnomes/images.php

[DeanWhitehouse]

It is , from your source code

    <td width="546"><img src="./uploads/1233033668.gif"  alt="Image can not be found" />   

   

   

    </td>

 

Near the bottom

[mike12255]

I just recreated the database and it appears to be working