Hi, I can't get this script to work.
I've used this exact script on other places and it works, but now i
get this error.
<codeWarning: mysql_fetch_array(): supplied argument is not a valid
MySQL result resource in C:\xampp\htdocs\uploads\login_script.php on
line 15 </code>
I can't see what is wrong.
Here is the script.
<code>
<?php
session_start();
$anvnamn = $_POST['usr'];
$losenord = $_POST['pwd'];
include "dbconnect.php";
$anv2 = mysql_real_escape_string($anvnamn, $dbconnect);
$los2 = mysql_real_escape_string($losenord, $dbconnect);
$sqlfraga = "SELECT anvnamn FROM administrator WHERE anvnamn = '" .
$anvnamn . "' AND losen = '" . $losenord . "'";
$res = mysql_query($sqlfraga, $dbconnect);
if($rad = mysql_fetch_array($res))
{
$_SESSION['logged_in_admin'] = true;
}
else
{
$_SESSION['logged_in_admin'] = false;
}
?>
<html>
<body>
<?php
if($_SESSION['logged_in_admin'])
{
echo("You are logged in");
include('index.php');
}
else
{
echo ("go away");
}
?>
</body>
</html>
</code>
6  1309 
On May 19, 1:36*pm, morph.1...@gmail.com wrote:
<codeWarning: mysql_fetch_array(): supplied argument is not a valid
MySQL result resource in C:\xampp\htdocs\uploads\login_script.php on
line 15 </code>
$res = mysql_query($sqlfraga, $dbconnect);
if($rad = mysql_fetch_array($res))
When the query fails, mysql_query() returns false, which results in
the error message you wrote. I am not sure if this is the case in your
situation, because this would also print a warning. Check the output
of mysql_query() and use mysql_error() to get the error message. mo********@gmail.com wrote:
Hi, I can't get this script to work.
I've used this exact script on other places and it works, but now i
get this error.
<codeWarning: mysql_fetch_array(): supplied argument is not a valid
MySQL result resource in C:\xampp\htdocs\uploads\login_script.php on
line 15 </code>
I can't see what is wrong.
Here is the script.
<code>
<?php
session_start();
$anvnamn = $_POST['usr'];
$losenord = $_POST['pwd'];
include "dbconnect.php";
$anv2 = mysql_real_escape_string($anvnamn, $dbconnect);
$los2 = mysql_real_escape_string($losenord, $dbconnect);
You create some escaped versions of the $_POST data...
$sqlfraga = "SELECT anvnamn FROM administrator WHERE anvnamn = '" .
$anvnamn . "' AND losen = '" . $losenord . "'";
.... but then fail to use them (SQL injection alert!).
$res = mysql_query($sqlfraga, $dbconnect);
Then fail to check whether $res is FALSE, which could be the case if
there was an issue with rights to the database.
if($rad = mysql_fetch_array($res))
Which would cause this to error as described.
So, the error said that $res wasn't valid, so why didn't you check what
was being used? Simple debugging...
Robin
On Mon, 19 May 2008 13:36:24 +0200, <mo********@gmail.comwrote:
Hi, I can't get this script to work.
I've used this exact script on other places and it works, but now i
get this error.
<codeWarning: mysql_fetch_array(): supplied argument is not a valid
MySQL result resource in C:\xampp\htdocs\uploads\login_script.php on
line 15 </code>
I can't see what is wrong.
Here is the script.
<code>
<?php
session_start();
$anvnamn = $_POST['usr'];
$losenord = $_POST['pwd'];
include "dbconnect.php";
$anv2 = mysql_real_escape_string($anvnamn, $dbconnect);
$los2 = mysql_real_escape_string($losenord, $dbconnect);
Proper escaping and then:
$sqlfraga = "SELECT anvnamn FROM administrator WHERE anvnamn = '" ..
$anvnamn . "' AND losen = '" . $losenord . "'";
.... using the unescaped variables!
You, my friend, are vulnerable to SQL injection. Use the $avn2 & $los2
variables in the query, that's why you escape()d them...
If you still have the same problem, echo $sqlfraga & mysql_error() to the
screen and check what's wrong with the query.
--
Rik Wasmus
....spamrun finished
On Mon, 19 May 2008 14:05:55 +0200, Robin <an**@somewhere.comwrote: mo********@gmail.com wrote:
>Hi, I can't get this script to work.
I've used this exact script on other places and it works, but now i
get this error.
<codeWarning: mysql_fetch_array(): supplied argument is not a valid
MySQL result resource in C:\xampp\htdocs\uploads\login_script.php on
line 15 </code>
I can't see what is wrong.
Here is the script.
<code>
<?php
session_start();
$anvnamn = $_POST['usr'];
$losenord = $_POST['pwd'];
include "dbconnect.php";
$anv2 = mysql_real_escape_string($anvnamn, $dbconnect);
$los2 = mysql_real_escape_string($losenord, $dbconnect);
You create some escaped versions of the $_POST data...
>$sqlfraga = "SELECT anvnamn FROM administrator WHERE anvnamn = '"
mo********@gmail.com escribió:
Hi, I can't get this script to work.
I've used this exact script on other places and it works, but now i
get this error.
<codeWarning: mysql_fetch_array(): supplied argument is not a valid
MySQL result resource in C:\xampp\htdocs\uploads\login_script.php on
line 15 </code>
I can't see what is wrong.
Speaking in plain English, this error message means that you can't fetch
rows from $res because the database query failed. So you need to check
whether the query fails or not:
$res = mysql_query($sqlfraga, $dbconnect);
if(!$res){
// Error: log it, abort or whatever
echo 'Query failed: ' . mysql_error();
}else{
// Read rows
}
I also recommend you to enable full error reporting (at least in your
dev box). Edit your php.ini file or add this to the top of the script:
ini_set('display_errors', 1);
error_reporting(E_ALL);
--
-- http://alvaro.es - Álvaro G. Vicario - Burgos, Spain
-- Mi sitio sobre programación web: http://bits.demogracia.com
-- Mi web de humor al baño María: http://www.demogracia.com
--
On May 19, 2:13 pm, "Álvaro G. Vicario"
<alvaroNOSPAMTHA...@demogracia.comwrote:
morph.1...@gmail.com escribió:
Hi, I can't get this script to work.
I've used this exact script on other places and it works, but now i
get this error.
<codeWarning: mysql_fetch_array(): supplied argument is not a valid
MySQL result resource in C:\xampp\htdocs\uploads\login_script.php on
line 15 </code>
I can't see what is wrong.
Speaking in plain English, this error message means that you can't fetch
rows from $res because the database query failed. So you need to check
whether the query fails or not:
$res = mysql_query($sqlfraga, $dbconnect);
if(!$res){
// Error: log it, abort or whatever
echo 'Query failed: ' . mysql_error();
}else{
// Read rows
}
I also recommend you to enable full error reporting (at least in your
dev box). Edit your php.ini file or add this to the top of the script:
ini_set('display_errors', 1);
error_reporting(E_ALL);
--
--http://alvaro.es- Álvaro G. Vicario - Burgos, Spain
-- Mi sitio sobre programación web:http://bits.demogracia.com
-- Mi web de humor al baño María:http://www.demogracia.com
--
tanks for the help all of you guys.. the escaping being wrong i was
already aware of, i was in a bit of hurry when i set them up and i saw
that it was wrong just after posting this...
anyways the problem was that i named the table administrators in the
database and i wrote administrator in the querry, so all i really
needed was an "s"... This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics | |
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