Just in case you were still looking for info on your question, I've put all
of the code from a page I have working that does pretty much what you're
looking for. I'm no guru, but the code is commented pretty well. The page
does a little more than what you have mentioned you're looking for, but
maybe some of the "extras" will be of use to you.
############################################################################
##
<?php
//This'll be a template page to be used as a basis for a content area
that's just open and waiting to be filled in with text or a custom table.
echo("
<div class='content'>
");
//set the directory. needs to be pulled in as a variable from a menu
choice or previous page
$dir = $incomingDirName;
//Handles the error in the case that the incoming directory can't be
found
for some reason.
if (! @openDir($dir))
{
echo ("<p class='body'>I'm sorry, but the image directory
".$incomingDirName." can not be found on this server.</p>");
exit();
}
/*
output the current directory information. testing purposes only.
$thisDir=getcwd();
echo "<p class='title'>Results of getcwd(): ".$thisDir."</p><br>\n";
echo "<p class='title'>Files in ".$dir.":</p><br>\n";
*/
//open up a table on the page
echo("
<table align='center' border='0' cellpadding='4px'>
<tr>
");
//scan the directory and load its contents into an array
$handle=opendir($dir);
while (false !== ($file = readdir($handle)))
{
$retVal[] = $file;
}
//close the directory read operation then sort the array
closedir($handle);
sort($retVal);
//return $retVal; //required if a Function
//initialize a counter for use in formatting the table
$counter = 0;
//iterate over array
for ($i = 0; $i < count($retVal); $i++)
{
//strip out the . and .. level directory information
if ($retVal[$i] != "." && $retVal[$i] != "..")
{
//look for directory contents that start with "tn_"
if (strstr($retVal[$i], "tn_"))
{
//if the file starts with tn_ set it for use
as the img src""
$thumImg = $retVal[$i];
//then strip off the tn_ and use the result as
the image to link to
$linkTo = substr_replace($retVal[$i], "", 0,
3);
//combine directory and file information into
one variable for ease of
coding
$fullThumPath = $dir.$retVal[$i];
$fullLinkToPath = $dir.$linkTo;
//counter starts at 0. for iterations 0-3, put
images into cells in
current row
if ($counter <= 3)
{
//drop the two variables into a line
to create table information
echo ("<td align='center'><a
href='".$fullLinkToPath."'
target='_top'><img src='".$fullThumPath."' alt='Thumbnail'
border='0'></a></td>");
$counter++;
}
//for iterations greater than 3 (four images)
close the current row,
start another and drop current image in first cell
else
{
echo ("</tr><tr><td align='center'><a
href='".$fullLinkToPath."'
target='_top'><img src='".$fullThumPath."' alt='Thumbnail'
border='0'></a></td>");
$counter = 1;
}
}
}
//This is the basis for checking for a directory without image
files.
It's a bit broken as it reports an error whenever a file that soesn't start
with "tn_" is encountered. Which is exactly what the if() statement tells it
to do. Needs a little work.
/*
if (! strstr($retVal[$i], "tn_"))
{
echo ("<p class='body'>I'm sorry, but this directory
does not contain
any thumbnail images. Please contact Rich and let him know he screwed
up!</p>");
}
*/
}
//ending image table tags
echo("
</tr>
</table>
");
//ending page div tag
echo("
</div>
");
?>
############################################################################
##
Rich
-----Original Message-----
From: markbm [mailto:[EMAIL PROTECTED]]
Sent: Saturday, July 06, 2002 7:35 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Retrieving/Displaying hyperlinked images with PHP
I am trying to build a "product detail" page that pulls data from a MYSQL
database using PHP. The data for the page includes product images, which I
am trying to link to (i.e. from their location on the web server) instead of
loading the images into the database. However, I cannot find any sample code
that seems to work. Two questions:
1. Is this possible (i.e. to store the HYPERLINK to the image in the
database , and as the results are returned to the product detail screen, the
image file will be displayed)? OR RATHER do I need to store the physical
image file in the database location and query it that way?
2. The code sample below contains several lines that show a field populated
with text that I am returning....the line under the //Test comment is the
field that I'm trying to pull an image back for:
printf("REL_PLAN7: %s<br>\n", mysql_result($result,0,"REL_PLAN7"));
printf("REL_PLAN8: %s<br>\n", mysql_result($result,0,"REL_PLAN8"));
printf("REL_PLAN9: %s<br>\n", mysql_result($result,0,"REL_PLAN9"));
//test
printf(mysql_result($result,0,<a href="FRONT_REND">FRONT_REND</a>);
NOTE: "FRONT_REND" is the name of the database field, and it contains a full
web address, not relative.
Any help would be GREATLY appreciated. Thanks.
Mark
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