getimagesize problem

dstonek

PHP 4.2.3
Thanks in advance for any help

<?php
if ($handle = opendir('C:/phpdev/www/img/')) {
while (false !== ($file = readdir($handle))) {
$size = getimagesize($file); //ERROR MESSAGE "unable to open for reading 'name of the file'"

}
closedir($handle);

}
?>

The following code is working fine

<?php
$file= "C:/phpdev/www/img/f1.jpg";
$size2 = getimagesize ($file);
echo "<img src= $file >" ;
echo $size2[3]; ?>

jnuneznyc

Add the path to the beginning of the $file

<?php
$path = 'C:/phpdev/www/img/';
if ($handle = opendir( $path )) { 
    while (false !== ($file = readdir($handle))) { 
        $size = getimagesize($path.$file);
    } 
    closedir($handle); 
} 
?>

dstonek

It works. Thank you very much, Daniel

dstonek

Here is the code I need.

Next, I manually save the list in a text file and then import it from MyAdmin to a MySQL table.
Is it possible to make this automatically with PHP? Thanks, Daniel.
PS: I have to create a form to enter the path.

<?php $path = "D:/...PATH.../";
if ($dir = opendir($path)) {
while (false !== ($file = readdir($dir))) {
if ($file != "." && $file != "..") {
$size = getimagesize($path.$file);
$width = $size[0];
$height = $size[1];
$opensize = $size[0]$size[1]3/1024/1024;
$path2 = $path.$file;
echo($path2.";".$width.";".$height.";".$opensize);
echo "</a><br />\n";

}
}
closedir($dir);
}
?>

jnuneznyc

Store the line in a variable then send it to MySQL with an INSERT statement...

if you need SQL help goto: http://www.sqlcourse.com/

use the mysql_query() function to send it to the DB