Home > Archive > PHP Language > November 2007 > Supplied argument is not a valid MySQL-Link resource
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| Author |
Supplied argument is not a valid MySQL-Link resource
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| j'ai ce message d'erreur "Supplied argument is not a valid MySQL-Link
resource "
ci-dessous mon code
<?
$connexion=mysql_connect("localhost","root","") or die(mysql_error());
$db=mysql_select_db("monsite",$connexion);
?>
<?php
$rep="vignettes";
$rep2="photos";
$id_rep = opendir($rep);
while ($fichier = readdir($id_rep))
{
$extension=substr($fichier,-3);
if($fichier!="." && $fichier!=".." && (eregi("gif",$extension)||
eregi("jpg",$extension)))
{
$nomimage = "$rep/$fichier";
$emplacement_fichier = $rep2 . "/" .substr("$fichier",1);
$nom_fichier= substr("$fichier",1,strpos("$fichier",".")-1);
//echo $nom_fichier;
//echo $nomimage;
$mavignette = addslashes(fread(fopen($nomimage, "rb"),
filesize($nomimage)));
// echo $mavignette ;
echo $emplacement_fichier;
//$sql="INSERT INTO photos (image_nom,image_vignette,image_fichier)
//VALUES ('$nom_fichier','mavignette','$emplaceme
nt_fichier')";
//$result=mysql_query($sql,$connexion) or die(mysql_error());
?>
<?php
$result=mysql_query("INSERT INTO photos
(image_nom,image_vignette,image_fichier)
VALUES ('$nom_fichier','$mavignette','$emplacem
ent_fichier')",
$connexion) or die(mysql_error());
}
}
mysql_close();
?>
merci d'avance
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| Olaf Schinkel 2007-11-19, 7:59 am |
| Hi!
> //$sql="INSERT INTO photos (image_nom,image_vignette,image_fichier)
> //VALUES ('$nom_fichier','mavignette','$emplaceme
nt_fichier')";
> //$result=mysql_query($sql,$connexion) or die(mysql_error());
What about a:
echo $sql;
Maybe wrong chars in it.
Post the whole errormessage here.
Olaf
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