adding to MySQL image from dir list, no reaction no error

Peuplarchie · 06-29-2010, 11:49 PM

Good day to you all,
I'm trying to read a directory and for each file add : path, name, ext, width and height to my database.

Right now the only thing I had succeed in to list the image directory.

I have tried out to add it to the database but no reaction, nothing is added to the db, no error ?

Here is my code:

PHP Code:


			
  <?php


// Connect to database


$con = mysql_connect("localhost","111",111");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("images", $con);


  
  $imgdir = 'Images/'; // the directory, where your images are stored
  $realimgdir = "Images/";
  $allowed_types = array('png','jpg','jpeg','gif','bmp'); // list of filetypes you want to show
  
  $dimg = opendir($imgdir);
  while($imgfile = readdir($dimg))
  {
   if(in_array(strtolower(substr($imgfile,-3)),$allowed_types))
   {
    $a_img[] = $imgfile;
    sort($a_img);
    reset ($a_img);
   }
  }
  
  $totimg = count($a_img); // total image number
   
  for($x=0; $x < $totimg; $x++)
  {
   $size = getimagesize($imgdir.'/'.$a_img[$x]);
  
   // do whatever
   $halfwidth = ceil($size[0]/2);
   $halfheight = ceil($size[1]/2);

$rid = $realimgdir;
$fname = substr($a_img[$x],0, -4);
$extens = substr($a_img[$x], -3);
$iwidth = $size[0];
$iheight = $size[1];

   
mysql_query("INSERT INTO gallery (id, path, name, ext, width, height)
VALUES ('$rid', '$fname', '$extens', '$iwidth', '$iheight'");  

   echo '<tr title="'.$a_img[$x].'" onMouseOver="this.className=\'highlight\'" onMouseOut="this.className=\'normal\'"></td><td>'.$realimgdir.'</td><td>'.substr($a_img[$x],0, -4).'</td><td style="text-align:center;"> '.substr($a_img[$x], -3).' </td><td style="text-align:center;"> '.$size[0].' </td><td style="text-align:center;"> '.$size[1].'</td></tr>';
  }
  
  ?>
</table>

Somebody have an idea

nonggiatu · 07-01-2010, 03:13 AM

You should catch the exception when working with DB and see if something is wrong. Maybe two reasons:
1. If the ID field is "auto increment", you should not specify the ID value to insert.
2. Is possible that there were two equal IDs ?