Prob script php : parse error
bonjour a tous je voudrais savoir ce que ve dire ce message d'erreur :
Parse error: parse error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING in c:\program files\easyphp1-7\www\calimero2\page\log.php on line 13
voici mon code :
<?php
/* Connexion et sélection de la base */
require_once("function.php");
$cnx = $_GET[connect()]; //connection a la base
$sql1 = "SELECT id_m, type_m FROM membre WHERE log_m = '$_GET['log']'";
$rst1 = send_sql($sql);
$name1 = mysql_fetch_array($rst1);
$id_log = $row[0];
$type = $row[1];
$sql2 = "SELECT id_m FROM membre WHERE pass_m = '$_GET['pass']'";
$rst2 = send_sql($sql2);
$name2 = mysql_fetch_array($rst2);
$id_pass = $row[0];
if($id_pass == 0)
{
$id_pass = a;
}
if($id_log == $id_pass)
{
include("$type");
}
else
{
include("index.php");
}
?>
merci
Parse error: parse error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING in c:\program files\easyphp1-7\www\calimero2\page\log.php on line 13
voici mon code :
<?php
/* Connexion et sélection de la base */
require_once("function.php");
$cnx = $_GET[connect()]; //connection a la base
$sql1 = "SELECT id_m, type_m FROM membre WHERE log_m = '$_GET['log']'";
$rst1 = send_sql($sql);
$name1 = mysql_fetch_array($rst1);
$id_log = $row[0];
$type = $row[1];
$sql2 = "SELECT id_m FROM membre WHERE pass_m = '$_GET['pass']'";
$rst2 = send_sql($sql2);
$name2 = mysql_fetch_array($rst2);
$id_pass = $row[0];
if($id_pass == 0)
{
$id_pass = a;
}
if($id_log == $id_pass)
{
include("$type");
}
else
{
include("index.php");
}
?>
merci
5 réponses
-
ModérateurPar sécurité, vérifie la validité de $type, MEME si ceci provient de ta BDD.
Johan
Gates gave you the windows.
GNU gave us the whole house.(Alexandrin) -
ModérateurEt là également:
$rst1 = send_sql($sql);
C'est $sql1 que tu voulais mettre je suppose.... -
Modérateur
johand@horus:~$ cat bidon.php <?php /* Connexion et sélection de la base */ require_once("function.php"); $cnx = $_GET[connect()]; //connection a la base $sql1 = "SELECT id_m, type_m FROM membre WHERE log_m = '$_GET['log']'"; $rst1 = send_sql($sql); $name1 = mysql_fetch_array($rst1); $id_log = $row[0]; $type = $row[1]; $sql2 = "SELECT id_m FROM membre WHERE pass_m = '$_GET['pass']'"; $rst2 = send_sql($sql2); $name2 = mysql_fetch_array($rst2); $id_pass = $row[0]; if($id_pass == 0) { $id_pass = a; } if($id_log == $id_pass) { include("$type"); } else { include("index.php"); } ?> johand@horus:~$ php -l bidon.php Parse error: parse error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING in bidon.php on line 9 Errors parsing bidon.php johand@horus:~$ vi bidon.php
Après éditionjohand@horus:~$ cat bidon.php <?php /* Connexion et sélection de la base */ require_once("function.php"); $cnx = $_GET[connect()]; //connection a la base $sql1 = sprintf("SELECT id_m, type_m FROM membre WHERE log_m = '%s'", $_GET["log"]); $rst1 = send_sql($sql); $name1 = mysql_fetch_array($rst1); $id_log = $row[0]; $type = $row[1]; $sql2 = sprintf("SELECT id_m FROM membre WHERE pass_m = '%s'",$_GET["pass"]); $rst2 = send_sql($sql2); $name2 = mysql_fetch_array($rst2); $id_pass = $row[0]; if($id_pass == 0) { $id_pass = a; } if($id_log == $id_pass) { include("$type"); } else { include("index.php"); } ?> johand@horus:~$ php -l bidon.php No syntax errors detected in bidon.php
Les messages d'erreur sont là pour être lus...
Ceci ne prédit en rien la validité fonctionnelle du code.
php -l fournit quelque oiptions utiles, non ?
Gates gave you the windows.
GNU gave us the whole house.(Alexandrin) -
ModérateurSalut,
L'erreur vient de là je pense:
$id_pass = a;
-
non je ne pense pas j'ai essayer de l'inicialiser a 999 mais c pareil
