What am I doing wrong with this login?

Sarahc · February 24, 2007

I am using the script below, but when I go to the next page it says I am not logged in, like it is not passing the login info. I posted the page that is saying I need to log in after the first script.

<?php

$login = $_REQUEST['login'];

$password = $_REQUEST['password'];

$sql = "select * from users where login = '$login' and password = '$password'";

$result = mysql_query($sql ,$db);

if ($myrow = mysql_fetch_array($result)) {

do {

$uid = $myrow["userid"];

$uname = $myrow["login"];

} while ($myrow = mysql_fetch_array($result));

$loggedin = true;

$upwd = $password;

$msg = "Welcome $uname, you are now logged in. <a href=index.php>Click here</a> to return to the front page.";

} else {

$loggedin = false;

$upwd = "";

$uid = "";

$uname = "";

$msg = "Sorry, that login and password is not valid. Click back on your browser to try sgain. If you have forgotten your password <a href=forgot.php>click here</a>. If you are a new user you will need to <a href=newuser.php>create a new login</a>";

}

session_register("loggedin");

session_register("upwd");

session_register("uid");

session_register("uname");

?>

<?php include("theader.php"); ?>

<tr>

<font face="Arial, Helvetica, sans-serif" size="3"><b>Add a Recipe</b></font>

</td></tr></table>

</td></tr>

<TR>

<?php

if (logincheck($uid, $upwd)) {

?>

</font>

<tr>

<label>If you are unsure, please do not select.</label>

</p>

<p>

<label>Gluten Free

yes</label>

</p>

<p>

<label>Casein Free

yes</label>

</p>

<p>

<label>Soy Free

yes</label>

</p>

<p>

<label>Yeast Free

yes</label>

</p>

<p>

<label>Phenol Free

yes</label>

<br>

</p><br>

Enter the title for your Recipe</th> </tr>

<tr>

</tr>

<tr>

<th colspan=2><font size=2 face='Arial, Helvetica, sans-serif'>Enter the Ingredients</font></th>

</tr>

<tr>

</tr>

<tr>

</tr>

<tr>

</tr>

<tr>

</tr>

<tr>

</tr>

<tr>

</tr>

<tr>

</tr>

<tr>

</tr>

<tr>

<th colspan=2><font size=2 face='Arial, Helvetica, sans-serif'>Enter the cooking directions</font></th>

</tr>

<tr>

</tr>

<tr>

<th colspan=2>Upload recipe picture</th>

</tr>

<tr>

</tr>

<tr>

<th colspan=2><font size=2 face='Arial, Helvetica, sans-serif'>Select a Categories</font></th>

</tr>

<?php

$sql = "select * from categories order by category";

$result = mysql_query($sql ,$db);

$newrowcount = 0;

printf("<table><tr>");

if ($myrow = mysql_fetch_array($result)) {

do {

if ($newrowcount == 4) {

$newrowcount = 0;

printf("</tr>");

printf("<tr>");

}

printf("<td><font size=2 face='Arial, Helvetica, sans-serif'><input type=radio name=catid value=%s> %s</font></td>", $myrow["categoryid"], $myrow["category"]);

$newrowcount++;

} while ($myrow = mysql_fetch_array($result));

}

for ($c = $newrowcount; $c < 4; $c++) {

printf("<td></td>");

}

printf("</tr></table>");

?>

</td></tr>

<tr>

<br>

</td>

</tr>

</table>

</form>

<?php

} else {

printf("<font size=2 face='Arial, Helvetica, sans-serif'><br>You need to be logged in before you can add a recipe</font>");

}

?>

</center>

</FONT>

</TD>

</TR>

</TABLE>

</td>

</tr>

</table>

</TD>

<?php include("tfooter.php"); ?>

mmarif4u · February 24, 2007

Where is ur session_start();

papaface · February 24, 2007

and use code tags!

magic2goodil · February 24, 2007

Also you may think about mysql_fetch_assoc() instead of mysql_fetch_array(), assoc fetches a less cmplex array that can still be accessed with the keys you are using.

Also, instead of:

if ($myrow = mysql_fetch_array($result)) {

         do {

            $uid = $myrow["userid"];
            $uname = $myrow["login"];

         } while ($myrow = mysql_fetch_array($result));

}

Perhaps think of something simpler in the future:

if ($myrow = mysql_fetch_array($result)) {
            
            while ($myrow = mysql_fetch_array($result)) {

                      $uid = $myrow["userid"];
                      $uname = $myrow["login"];

            }

}

I just think the do() is unnecessary.

Sarahc · February 24, 2007

None of these suggestions seem to help. Any more?

magic2goodil · February 24, 2007

Instead of:

if ($myrow = mysql_fetch_array($result)) {

         do {

            $uid = $myrow["userid"];
            $uname = $myrow["login"];

         } while ($myrow = mysql_fetch_array($result));

         $loggedin = true;
         $upwd = $password;
         $msg = "Welcome $uname, you are now logged in.  <a href=index.php>Click here[/url] to return to the front page.";

      }

Try this:


$myrow = mysql_fetch_assoc($result) or die(mysql_error());
$affected_rows = mysql_num_rows($result) or die(mysql_error());

if ($affected_rows == 1) {

            $uid = $myrow["userid"];
            $uname = $myrow["login"];

            $loggedin = true;
            $upwd = $password;
            $msg = "Welcome $uname, you are now logged in.  <a href=index.php>Click here[/url] to return   to the front page.";

            session_register("loggedin");
            session_register("upwd");
            session_register("uid");
            session_register("uname");

}

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What am I doing wrong with this login?

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