Dynamic thumbnails

Report · Jun 22, 2006

I am relatively new to PHP and I am working on a project that uses various
sizes of pictures and I want to create a thumbnails of these image. The
references to these images are held in a database and I have found some code
that uses getimagesize to get the size of the image I can get this to work
if I reference and image directly my problem is I'm having a difficult time
understanding the PHP code to pull the image from the database. I have tried
various methods with the current code that pulls images for display but I
keep getting syntax errors and I really do not know what to do differently
below is a copy of the code than I am using the code was developed in dream
weaver for accessing the database. Any help would be greatly appreciated.

The error I get is Line 76 is the line with the pointer to the image name
in the database.

Parse error: parse error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting
T_STRING or T_VARIABLE or T_NUM_STRING in
C:\Accounts\antiques\wwwRoot\TestSite\Untitled-2.php on line 76

I am sure that it is probably something really stupid that I am over
looking

Thanks for your help

<body>
<img src="../AAAIMAGES/<?php echo $row_Recordset1['Image']; ?>.jpg"
width="235" height="225" /> This displays the mage from the DB
 
 
 

<img src="../AAAIMAGES/b151.jpg" /> This displays the image in thumbnail
when I point to the image directly
<?php
$image = "..\AAAIMAGES\b151.jpg";
echo $image;
$size = getimagesize("$image");
$height = $size[1];
$width = $size[0];
if ($height > 150)
{
$height = 200;
$percent = ($size[1] / $height);
$width = ($size[0] / $percent);
}
else if ($width > 150)
{
$width = 200;
$percent = ($size[0] / $width);
$height = ($size[1] / $percent);
}
echo "<img src =\"../AAAIMAGES/b151.jpg\" height=\"$height\"
width=\"$width\" />";
?>

<img src="../AAAIMAGES/b151.jpg" /> This displays the image
<?php

// line 76 is below

$image = "../AAAIMAGES/<?php echo $row_Recordset1['Image']; ?>.jpg";
echo $image;
$size = getimagesize("$image");
$height = $size[1];
$width = $size[0];
if ($height > 150)
{
$height = 200;
$percent = ($size[1] / $height);
$width = ($size[0] / $percent);
}
else if ($width > 150)
{
$width = 200;
$percent = ($size[0] / $width);
$height = ($size[1] / $percent);
}
echo "<img src ="../AAAIMAGES/<?php echo $row_Recordset1['Image'];
?>.jpg" height=\"$height\" width=\"$width\" />";
?>

Report · Jun 23, 2006

TAD wrote:
> Parse error: parse error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting
> T_STRING or T_VARIABLE or T_NUM_STRING in
> C:\Accounts\antiques\wwwRoot\TestSite\Untitled-2.php on line 76

I haven't checked the rest of your code but the parse error is caused by
nesting PHP tags within a PHP block. PHP blocks cannot be nested. The
following line

$image = "../AAAIMAGES/<?php echo $row_Recordset1['Image']; ?>.jpg";

should be

$image = "../AAAIMAGES/{$row_Recordset1['Image']}.jpg";

You need to surround $row_Recordset1['Image'] by curly braces to include
this sort of array variable in a double-quoted string.

--
David Powers
Author, "Foundation PHP for Dreamweaver 8" (friends of ED)
Author, "Foundation PHP 5 for Flash" (friends of ED)
http://foundationphp.com/