i think the sql insert statement should be:
insert into images(sixfourdata) values ('$encoded');
Rui Cunha
Alessandro Folghera writes:
Anybody has an idea about what happened to me ?
The following script should just charge in a mysql db the images located in
a directory. Unfortunately everytime I call "readdir.php" (never mind if
there are or not new images) mysql is charging 2 or 6 copies of the same
last image uploaded into the directory. Anybody may explain to me where I'm
failing?
Thanks for all the phpers!
Alessandro
<?
$dbcnx = mysql_connect("localhost", "root", "password"); mysql_select_db("news");
if (!$dbcnx) { echo( "<p>connection to database server failed!</p>");
exit();
}
if (! @mysql_select_db("news") )
{ echo( "<p>Image Database Not Available!</p >" ); exit(); }
$path = "./";
$dir_handle = @opendir($path) or die("Unable to open directory $path");
while ($file = readdir($dir_handle)) {
$filetyp = substr($file, -3);
if ($filetyp == 'gif' OR $filetyp == 'jpg') {
$handle = fopen($path . "/" . $file,'r');
$file_content = fread($handle,filesize($path . "/" . $file));
fclose($handle);
$encoded = chunk_split(base64_encode($file_content)); $sql = "INSERT INTO images SET sixfourdata='$encoded'";
@mysql_query($sql);
}
}
closedir($dir_handle);
echo("complete");
?>
*************************************
* Rui Pedro Cunha *
* Dpto. de Ciências e Tecnologias *
* Universidade Autónoma de Lisboa *
* Rua de Santa Marta, 56, *
* 1169-023 Lisboa *
* Telefone (+351) 21 317 76 35/49 *
* Fax (+351) 21 353 37 02 *
* Url : http://www.ual.pt/dct/ *
* E-mail: rui.cunha@xxxxxx *
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