php Login script issue

[Chris Carter]

Hi,

Its just a login and password validation that I am trying to achieve. If the username is correct then the person is able to view certain page, if incorrect then he is directed elsewhere.

<?
$userid=mysql_real_escape_string($userid);
$password=mysql_real_escape_string($password);

if($rec=mysql_fetch_array(mysql_query("SELECT * FROM tablename WHERE userName='$userName' AND password = '$password'"))){
        if(($rec['userName']==$userName)&&($rec['password']==$password)){
         include "../include/newsession.php";
            echo "<p class=data> <center>Successfully,Logged in<br><br> Log OUT <br><br>Click here if your browser is not redirecting automatically or you don't want to wait.<br></center>";
     print "<script>";
       print " self.location='submit-store-details.php';"; // Comment this line if you don't want to redirect
          print "</script>";

                                }
                }
        else {

                session_unset();
echo "Wrong Login. Use your correct  Userid and Password and Try <br><center><input type='button' value='Retry' onClick='history.go(-1)'></center>";
               
        }
?>

I am getting this error when I am using this code:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in thispage.php on line 37
Wrong Login. Use your correct Userid and Password and Try

Why does it show up everytime and whats wrong with mysql_fetch_array().

Please advice also if there is some other way available please help me try that.

Thanks,

Chris

On 9/16/07, Chris Carter <chandan9sharma@...> wrote:
>
> Hi,
>
> Its just a login and password validation that I am trying to achieve. If the
> username is correct then the person is able to view certain page, if
> incorrect then he is directed elsewhere.
>
> <?
> $userid=mysql_real_escape_string($userid);

Here you call it $userid

> $password=mysql_real_escape_string($password);
>
> if($rec=mysql_fetch_array(mysql_query("SELECT * FROM tablename WHERE
> userName='$userName' AND password = '$password'"))){

and here you call it $userName. If this is the full code, $userName is
not set here, and it would result in query userName='' and mysql_query
will return FALSE, which isn't a valid mysql resource for
mysql_fetch_array.

>        if(($rec['userName']==$userName)&&($rec['password']==$password)){
>         include "../include/newsession.php";
>            echo "<p class=data> <center>Successfully,Logged in<br><br>
> logout.php  Log OUT  <br><br> welcome.php Click here if your browser is not
> redirecting automatically or you don't want to wait. <br></center>";
>     print "<script>";
>       print " self.location='submit-store-details.php';"; // Comment this
> line if you don't want to redirect
>          print "</script>";
>
>                                }
>                }
>        else {
>
>                session_unset();
> echo "Wrong Login. Use your correct  Userid and Password and Try
> <br><center><input type='button' value='Retry'
> onClick='history.go(-1)'></center>";
>
>        }
> ?>
>
> I am getting this error when I am using this code:
>
> Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result
> resource in thispage.php on line 37
> Wrong Login. Use your correct Userid and Password and Try
>
> Why does it show up everytime and whats wrong with mysql_fetch_array().
>
> Please advice also if there is some other way available please help me try
> that.
>
> Thanks,
>
> Chris

I advice you to split the code up in 2 seperate actions, and check for errors.

> if($rec=mysql_fetch_array(mysql_query("SELECT * FROM tablename WHERE userName='$userName' AND password = '$password'"))){

would become:
$result = mysql_query("SELECT * FROM tablename WHERE
userName='$userName' AND password = '$password'") or die
(mysql_error());
// You could also add some checks here with mysql_num_rows for example...
if($rec=mysql_fetch_array($result)){

Tijnema


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[snsanju]

Hi,

$result = mysql_query("SELECT * FROM tablename WHERE
userName='$userName' AND password = '$password'");
if($rec = mysql_fetch_array($result)){
        //your code
}

Try like this it may solve. It may solve your problem
Don't try to fetch the result from one single line code.

Warm Regards,
Sanjeev
http://www.sanchanworld.com/
http://webdirectory.sanchanworld.com - Submit your website URL
http://webhosting.sanchanworld.com - Choose your best web hosting plan
-----Original Message-----
From: Chris Carter [mailto:chandan9sharma@...]
Sent: Sunday, September 16, 2007 3:10 PM
To: php-general@...
Subject: [PHP] php Login script issue


Hi,

Its just a login and password validation that I am trying to achieve. If the
username is correct then the person is able to view certain page, if
incorrect then he is directed elsewhere.

<?
$userid=mysql_real_escape_string($userid);
$password=mysql_real_escape_string($password);

if($rec=mysql_fetch_array(mysql_query("SELECT * FROM tablename WHERE
userName='$userName' AND password = '$password'"))){
        if(($rec['userName']==$userName)&&($rec['password']==$password)){
         include "../include/newsession.php";
            echo "<p class=data> <center>Successfully,Logged in<br><br>
logout.php  Log OUT  <br><br> welcome.php Click here if your browser is not
redirecting automatically or you don't want to wait. <br></center>";
     print "<script>";
       print " self.location='submit-store-details.php';"; // Comment this
line if you don't want to redirect
          print "</script>";

                                }
                }
        else {

                session_unset();
echo "Wrong Login. Use your correct  Userid and Password and Try
<br><center><input type='button' value='Retry'
onClick='history.go(-1)'></center>";
               
        }
?>

I am getting this error when I am using this code:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result
resource in thispage.php on line 37
Wrong Login. Use your correct Userid and Password and Try

Why does it show up everytime and whats wrong with mysql_fetch_array().

Please advice also if there is some other way available please help me try
that.

Thanks,

Chris
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Sent from the PHP - General mailing list archive at Nabble.com.

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[Bastien Koert]

argh! hotmail sucks

I don't see in the script where you are using $_POST / $_GET / $_REQUEST to access tha data from the form. Its likely that the example you are following is old and uses 'register_globals'. Since register_globals is a huge security hole and is not active in any new installations of PHP you need to change your script to use the above methods to get the form data. The error you are getting is due to the fact that you are not passing in the values to the sql and not getting a valid result

Note that the below example fixes your issue but DOES NOT do any validation, which you really should do before passing your data to the sql...




$userid=mysql_real_escape_string($_POST['userid']);  
$password=mysql_real_escape_string($_POST['password']);

if($rec=mysql_fetch_array(mysql_query("SELECT * FROM tablename WHERE> userName='$userName' AND password = '$password'"))){

 if(($rec['userName']==$userName)&&($rec['password']==$password))


bastien



----------------------------------------> Date: Sun, 16 Sep 2007 02:39:57 -0700> From: chandan9sharma@...> To: php-general@...> Subject: [PHP] php Login script issue>>> Hi,>> Its just a login and password validation that I am trying to achieve. If the> username is correct then the person is able to view certain page, if> incorrect then he is directed elsewhere.>>  $userid=mysql_real_escape_string($userid);> $password=mysql_real_escape_string($password);>> if($rec=mysql_fetch_array(mysql_query("SELECT * FROM tablename WHERE> userName='$userName' AND password = '$password'"))){> if(($rec['userName']==$userName)&&($rec['password']==$password)){> include "../include/newsession.php";> echo " Successfully,Logged in> logout.php Log OUT  welcome.php Click here if your browser is not> redirecting automatically or you don't want to wait. ";> print "";>> }> }> else {>> session_unset();> echo "Wrong Login. Use your correct Userid and Password and Try>  onClick='history.go(-1)'>";>> }> ?>>> I am getting this error when I am using this code:>> Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result> resource in thispage.php on line 37> Wrong Login. Use your correct Userid and Password and Try>> Why does it show up everytime and whats wrong with mysql_fetch_array().>> Please advice also if there is some other way available please help me try> that.>> Thanks,>> Chris> --> View this message in context: http://www.nabble.com/php-Login-script-issue-tf4450691.html#a12698139> Sent from the PHP - General mailing list archive at Nabble.com.>> --> PHP General Mailing List (http://www.php.net/)> To unsubscribe, visit: http://www.php.net/unsub.php>

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