My image wont display

php_beginner_83

Hi All

I'm trying to create my own photo gallery. I'm still new at php and am slowly trying to build it up. My problem at the minute is that my image wont display. What I get is the box with the red cross in the middle.
My code is...

global $photos, $dir;

$galleryLayout =  array();

// first pic.
$currentPic = $dir . $photos[0];

//build html layout for the gallery.  using a table
$galleryLayout[] = '<table width="100%">';
$galleryLayout[] = '<tr><td width="33%">Previous Link will go here</td>';
$galleryLayout[] = '<td width="33%" align="center"></td>';
$galleryLayout[] = '<td align="right" width="33%">Next Link will go here</td></tr>';
$galleryLayout[] = '<tr><td align="center"><img src =' . $currentPic . '/>image goes here</td></tr>';
$galleryLayout[] = '</table>';


foreach ($galleryLayout as $value)
{
    echo $value;
}

Thanks

coldwerturkey

then the url of the image is wrong. fix it, & the image will show.

also that's an ......interesting....... way to make and print a string.

how about just

$galleryLayout = '<table width="100%">';
$galleryLayout .= '<tr><td width="33%">Previous Link will go here</td>';
//etc

echo $galleryLayout;

php_beginner_83

Well like I said, I'm a beginner and pretty much learning by trial and error.

bradgrafelman

Even simpler:

echo '<table width="100%">
<tr><td width="33%">Previous Link will go here</td>
<td width="33%" align="center"></td>
<td align="right" width="33%">Next Link will go here</td></tr>
<tr><td align="center"><img src="' . $currentPic . '"/>image goes here</td></tr>
</table>';

or:

<table width="100%">
<tr><td width="33%">Previous Link will go here</td>
<td width="33%" align="center"></td>
<td align="right" width="33%">Next Link will go here</td></tr>
<tr><td align="center"><img src="<?php echo $currentPic ?>"/>image goes here</td></tr>
</table>

Also note that in my above code snippet I fixed an HTML error in your code; attribute values should always be quoted.

php_beginner_83

Thanks Bradfagelman!
I have a problem though. I used your suggestion but now my variable $currentPic is empty when I used it your way. $currentPic is first initialised in a function within <?php ?> tags earlier in the script. Outside of these tags I used html for the table and then include the "<? php echo $currentPic ?>" like you suggested and it's coming up empty.
Any ideas what I'm doing wrong??

Thanks :-D I really appreciate your help.

bradgrafelman

Can you show us the code you're using now? Also, in the code snippet you first posted, you never showed where $dir or $photos come from.

php_beginner_83

Hey

I'm been messing around with my code and finally have something working, but I'd like to ask your opinion on something. So I'm trying to create a photo gallery and so far I have 2 links, Previous and Next. Clicking on these will navigate through the pictures. However, at the minute I only have it displaying one predefined picture when you click 'Previous' and another when you click 'Next'. I want to ask your opinion on which way or how would be the best way to loop through the pictures or determine which one is being displayed and then get the next one or previous one in the sequence, depending on which link the user clicks.
My code so far is..

<?php
// directory where photos are kept
$dir = "images/thumbnails/";
$photos = array();

if($handle = opendir($dir))
{
while(false !== ($file = readdir($handle)))
{
if ($file != "." && $file != ".." )
{
$photos[] = $file;
}
}
closedir($handle);
}

// count the no of photos
$noOfPics = count($photos);
$prev = 2;
$next = 4;
echo '<a href=My_Try_At_Album.php?p=' . $prev . '>Previous!!</a>';
echo '</br></br><a href=My_Try_At_Album.php?p=' . $next . '>Next!!</a>';

if (isset($GET['p']))
{
if (is_numeric($GET['p']))
{
$picIndex = $_GET['p'];
}
else
{
$picIndex = 1;
}

$currentPic = $photos[$picIndex];

echo '</br></br><img src="' . $dir . $currentPic .'"/>';

}

Thanks again :-)