fixing error message for session id

rdkd1970 · May 6, 2011

I am so glad someone told me about putting in the error_reporting ALL because I was not sure why my id for members was not picking up on the next pages. Can someone let me know how I can fix this I thought it automatically picks it up and sets it to private pages for each member.

Notice: Undefined index: id in /home/ebermy5/public_html/login.php on line 25

<?php

session_id();

session_start();

?>

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">

<head>

<title>Welcome</title>

</head>

<body>

<?php

/* Program: New_member.php

* Desc: Displays the new member welcome page. Greets

* member by name and gives a choice to enter

* restricted section or go back to main page.

*/

error_reporting(E_ALL);

ini_set("display_errors", 1);

$firstname = '';

include('Connections/connect_to_mysql.php');

$result = mysql_query("SELECT firstname FROM `Members` WHERE id='{$_SESSION['id']}'");

$row = mysql_fetch_array($result);

if ($firstname == ''){ //condition, is name equal to lower case firstname notice we use == and not =

echo "Welcome, $firstname";

} else { //so incase the condition is not as expected

echo "Sorry you are not $firstname";

}

?>

<p>Your new Member accounts lets you enter the members only section

of our web site. You'll find special discounts, a profile of matches,

live advise from experts, and much more.</p>

<p>Your new Member ID and password were emailed to you. Store them

carefully for future use.</p>

Glad you could join us!</p>

<input type="submit"

value="Enter the Members Only Section">

</form>

</form>

</div>

</body>

</html>

denno020 · May 6, 2011

session_start() needs to be the very first thing on your page. Move it about session_id() and see what happens..

Denno

rdkd1970 · May 6, 2011

I get the same message. Could it be my form. I see it is downloaded to my phpmyadmin.

denno020 · May 6, 2011

I'm not familiar with the use of the curly braces around $_SESSION['id'].. do you have a specific reason for this? This is the line that the error is appearing on, so I would try removing the curly braces, leave the line reading like this:

$result = mysql_query("SELECT firstname FROM `Members` WHERE id=$_SESSION['id']");

Denno

rdkd1970 · May 6, 2011

I pasted this over the one I have and it just states in the codes there is a syntax error on that line.

denno020 · May 6, 2011

What you could try doing is assigning $_SESSION['id'] to a variable and then using that variable in the where clause. I have a feeling there is issues when it comes to accessing the session variable directly in the where clause of SQL... I think I've heard of this before..

Denno

rdkd1970 · May 6, 2011

Okay I tried it as follows

$_SESSION='id';

and in the $result = $query i put where $_SESSION='id'

it added to db and it does not show error message but it is not echoing the Welcome, firstname. It shows Welcome,

denno020 · May 6, 2011

you should have done

$id = $_SESSION['id'];

Then put $id in the query..

Denno

rdkd1970 · May 6, 2011

I know it is something so easy that we are missing but me being new to this I would not have a clue. Here is my codes and then the message I got this time

error_reporting(E_ALL);

ini_set("display_errors", 1);

$firstname = '';

$id = $_SESSION['id'];

include('Connections/connect_to_mysql.php');

$result = mysql_query("SELECT firstname FROM `Members` WHERE $id=$_SESSION");

$row = mysql_fetch_array($result);

Notice: Undefined index: id in /home/ebermy5/public_html/login.php on line 23

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/ebermy5/public_html/login.php on line 27

Welcome,

denno020 · May 6, 2011

This is how your SQL statement should read

SELECT firstname FROM `Members` WHERE database_id=$id

database_id will be the id inside your database, obviously.. So you're checking those values for the value contained inside the session id variable, which you have saved into it's very own variable called $id. the $ sign before the name indicates it's a php variable also.. In case you didn't know .

Try that and let me know how you go.

Denno

rdkd1970 · May 6, 2011

Same result with error message. :'(

denno020 · May 6, 2011

Please post your whole php file so I can look through it. Also, be sure to post them between the php tags, by doing:

[ p h p ]

//your code goes here

[ / p h p ]

(without any of the spaces)

rdkd1970 · May 6, 2011

form

<?php

@include_once ("Connections/connect_to_mysql.php");

$err='';

if($_POST["submit"]){

// Validate form data

if($_POST["firstname"]=='') $err.='Please enter First Name<br>';

if($_POST["email"]=='') $err.='Please enter Email<br>';

if($err==''){

// Check if there are duplicate entries in the 'contacts' table

$results = mysql_query("SELECT id FROM `Members` WHERE firstname='".addslashes($_POST["firstname"])."' and Email='".addslashes($_POST["email"])."'");

if($row = mysql_fetch_array($results)){

$err.='Can not add duplicate entry<br>';

}

else{

// adding new record to 'contacts' table

mysql_query("INSERT INTO Members (firstname,lastname,country,Email)

values ('".addslashes($_POST["firstname"])."','".addslashes($_POST["lastname"])."','".addslashes($_POST["country"])."','".addslashes($_POST["email"])."')");

// redirecting to success screen

if($results){

header("Location: login.php");

}else

die(mysql_error());

}

?>

<html>

<head>

<title>Add New Contact</title>

</head>

<body>

<h2>Register with us</h2>

<?php echo $err==''?'''<p style="color:red;">'.$err.'</p>') ?>

<tr>

<td valign="middle">First Name:</td>

</tr>

<tr>

</tr>

<tr>

<td valign="middle">Country:</td>

</tr>

<tr>

<td valign="middle">Email:</td>

</tr>

</table><br>

</form>

</body>

</html>

Welcome page (login.php)

<?php

session_start();

session_id();

?>

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">

<head>

<title>Welcome</title>

</head>

<body>

<?php

/* Program: New_member.php

* Desc: Displays the new member welcome page. Greets

* member by name and gives a choice to enter

* restricted section or go back to main page.

*/

error_reporting(E_ALL);

ini_set("display_errors", 1);

$firstname = '';

$database_id = '';

include('Connections/connect_to_mysql.php');

$result = mysql_query("SELECT firstname FROM `Members` WHERE database_id=$id");

$row = mysql_fetch_array($result);

if ($firstname == ''){ //condition, is name equal to lower case firstname notice we use == and not =

echo "Welcome, $firstname";

} else { //so incase the condition is not as expected

echo "Sorry you are not $firstname";

}

?>

<p>Your new Member accounts lets you enter the members only section

of our web site. You'll find special discounts, a profile of matches,

live advise from experts, and much more.</p>

<p>Your new Member ID and password were emailed to you. Store them

carefully for future use.</p>

Glad you could join us!</p>

<input type="submit"

value="Enter the Members Only Section">

</form>

</form>

</div>

</body>

</html>

denno020 · May 6, 2011

Ok... So when I put 'database_id', I was kinda hoping you'd get the drift of what I was suggesting, and change the name to match whatever it is in your database. What is the name of the id field in your database? This is what you need to put where I suggested 'database_id'..

You also didn't assign anything to the php variable $id... You need to have $id = $_SESSION['id']; somewhere before your query.. Put it just below where you've got $firstname = ' ';

Once you have made those changes, let me know how it works.

Denno

rdkd1970 · May 6, 2011

I had the id in my db. I will try fix this to your suggestions and get back to you.

rdkd1970 · May 6, 2011

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/ebermy5/public_html/login.php on line 27

Welcome,

This is the corrections I made $id=''; and in the select firstname from members.... line so I think it is now the $row not getting the results or I did the result wrong.

error_reporting(E_ALL);

ini_set("display_errors", 1);

$firstname = '';

$id = '';

include('Connections/connect_to_mysql.php');

$result = mysql_query("SELECT firstname FROM `Members` WHERE firstname=$id");

$row = mysql_fetch_array($result);

denno020 · May 6, 2011

I just have to let you know that it's past 4am here, and I'm going to bed.. When I wake tomorrow, I will look at your problem some more and offer some more suggestions (of which I already have a few, but I want sleep). From the looks of it, its homework of some sort, so I'll let you continue to work on it and see if you can't fix it yourself. I would suggest youtube videos and Google.

Some things to think about: where does the information come from the user that is then used to query the database? (so where does the user put in their username and password?)

That's all for tonight. I should post back tomorrow at some point, if not then the day after..

Denno

rdkd1970 · May 6, 2011

I am getting the same error message with the row=results.

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fixing error message for session id

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