how to display pictures

i have a folder on my server with pics in it and i want to be able to display them on a web page and havnt got a clue where to start
any info please
Doug

 

you can display it by below code
<img src="folder/abc.jpg"> or <img src="folder/abc.jpg" width='500' height='500'>

 

with this code i have to write in all names
is there a way to dislay all without having to name them all in code
cheers
Doug

 

ok this is what i have so far puts pics on page but dont line up very well
and would like a name displayed with each pic

//Outputs the image and other data

Echo " ".$info['pet'] . " <br>";
Echo "<img src=http://www.hire.co.uk/pics/".$info['pic'] ." alt=\"Image\" align=\"left\" width=\"200px\" height=\"150px\" hspace=\"10px\" vspace=\"8px\"> <br/>";

PHP:

cheers
Doug

 

You browse the directory (using dir function) to get all file names, and after display them on your web.

 

Here's something I wrote quickly to show images in a directory in a tabular format.

Note that it does not work recursively, that is, it will not traverse sub-directories.

<table>
  <tr>
<?php
// Path to images
$sPath = './';
// Number to show per row
$iNum = 3;
// Allowed images
$aImages = array('jpg', 'jpeg', 'gif', 'png');

// Open the directory
$rDir = opendir($sPath);

// Initalise a counter for number of images
$iCount = 0;

// Loop over directory
while( ($sFile = readdir($rDir)) !== false )
{
    // Add path to filename
    $sFile = $sPath . $sFile;
    // Get file extension
    $sExt  = strtolower(pathinfo($sFile, PATHINFO_EXTENSION));

    // If it is an image...
    if ( in_array($sExt, $aImages) )
    {
        // Increase counter and check if it is neccessary to start a new row
        if ( $iCount++ % $iNum == 0 && $iCount > 1 )
        {
            echo '  </tr>' . "\n\n";
            echo '  <tr>' . "\n";
        }     
     
        // ...add image into table cell
        echo '    <td><img src="' . $sFile . '"></td>' . "\n";
    }
}

// If there are no images...
if ( $iCount == 0 )
{
    // ...output a message saying so
    echo '    <td>No images to display</td>' . "\n";   
}
else
{
    // ...otherwise complete the table by adding blank cells
    
    // Calculate how many blank cells are required
    $iLeft = $iNum - ($iCount % $iNum);
    // If anything other than a whole row is required...
    if ( $iLeft < $iNum )
    {
        // ...add the blank cells as many times as neccessary
        for($i = 0; $i < $iLeft; ++$i)
        {
            echo '    <td>&nbsp;</td>' . "\n";
        }
    }
}
?>
  </tr>
</table>

PHP:

 

exec('ls /dirname', $result);

 

Have you considered an open-source solution? There are MANY gallery scripts available. I use ImageView @ http://www.blackdot.be/?inc=projects/imageview.

 

how they're displayed will depend on how you use HTML basic design code
in the echo statement where it shows <br> ..thats a page break or (next line) so all your images are probably stacked vertically. If its displaying your images and you don't like how it displays them , then add HTML code to the echo statement. and be careful of double quotations so you dont get errors.