Undefined Offset Notice Warning

[Glese]

Here is an example of the beginning of a rating system script:

 

// RATING SYSTEM - START

if ((isset($array[0]) || isset($array[1])) && isset($_SESSION['user_id'])) {

// check to see if user has already voted on given contribution
$query = "SELECT * FROM votes WHERE con_id = '$array[2]' AND user_id = '$array[3]'";
$query_run = mysqli_query ($dbc, $query) or die (mysqli_error($dbc)); 
$num_rows = mysqli_num_rows ($query_run);
$assoc = mysqli_fetch_assoc ($query_run);
$rating = $assoc['rating'];


// check to see if it's user's own contribution
$query = "SELECT * FROM con WHERE con_id = '$array[2]'";
$query_run = mysqli_query ($dbc, $query) or die (mysqli_error ($dbc));
$assoc = mysqli_fetch_assoc ($query_run);
$con_user_id = $assoc['user_id'];

 

 

For every array variable I am getting the error message "Undefined Offset".

 

Any ideas how to solve this one, I do not fully know yet what is common practice in these cases. The warnings also do not always happen, they usually happen, they happen in a specific moment when I use a drop down menu, to show table content.

 

[KevinM1]

Undefined offset means you don't have a value in array[n], where n is the offset.

[Glese]

Excuse my confusion, I am a new comer, so how would I avoid the notice, obviously I am getting the notice, because the function is not in use, thus the variable is empty, any ideas?

[KevinM1]

Well, where does $array come from?

[ManiacDan]

if ((isset($array[0]) || isset($array[1])) && isset($_SESSION['user_id'])) {

// check to see if user has already voted on given contribution
$query = "SELECT * FROM votes WHERE con_id = '$array[2]' AND user_id = '$array[3]'";

That section checks for array elements zero and one, and then uses array elements two and three.  If we had the actual error message, we might be able to tell if that's the problem.

[Glese]

This is the actual error message:

Notice: Undefined offset: 2 in C:\xampp\htdocs\php_projects\myproject\controller\rating_system\rating_system_func.php on line 14

Notice: Undefined offset: 3 in C:\xampp\htdocs\php_projects\myproject\controller\rating_system\rating_system_func.php on line 14

Notice: Undefined offset: 2 in C:\xampp\htdocs\php_projects\myproject\controller\rating_system\rating_system_func.php on line 22

Notice: Undefined offset: 2 in C:\xampp\htdocs\php_projects\myproject\controller\rating_system\rating_system_func.php on line 64

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1[b][/b]

[KevinM1]

That's telling you, and confirms Dan's suspicions, that your $array only has two elements - $array[0] and $array[1].

 

So, again, where does $array come from?  What would lead you to believe it would have additional elements?

[Glese]

Array comes from this function, you can see array at the very bottom:

function knuffix_list ($query, $dbc) {

$data = mysqli_query ($dbc, $query) or die (mysqli_error ($dbc));						

// Loop through the array of data
while ($row = mysqli_fetch_array ($data)) {

	global $array;
                

                

	// Variables for the table
	$con_id = $row['con_id'];
	$likes_count = $row['likes'];
	$dislikes_count = $row['dislikes'];
	$dbuser_name = $row['nickname'];
	$user_id = $row['user_id'];
	$contribution = $row['contribution'];	
	$title = $row['name'];
                
                $avatar_file_name = $row['avatar'];
                $avatar_folder = "../profile/avatar/";

               
                  
                
                
                
                // The TABLE

                include('character_artwork_table.php');
                    
                
   
                
                
                
                
                // POST BUTTONS inside the table
                if (isset($_POST['likes']))
	$likes = $_POST['likes'];
                if (isset($_POST['dislikes']))
	$dislikes = $_POST['dislikes'];
                if (isset($_POST['hidden_con_id'])) {
	$con_id = $_POST['hidden_con_id'];
	//$favorite = $_POST['favorite'];
                }
                
               
	// $array = array ($likes, $dislikes, $con_id, $user_id);
                
                if (isset($likes)) 
                    $array[] = $likes;
                if (isset($dislikes))
                    $array[] = $dislikes;
                if (isset($con_id))
                    $array[] = $con_id;
                if (isset($user_id)) {
                    $array[] = $user_id;
                }

}

mysqli_close($dbc);

}

[ManiacDan]

So at least two of those are unset.  See how that works?  Your array is randomly one or more of those 4 items, yet your function assumes all 4 will be there.

[KevinM1]

Also, slight digression, but never use the 'global' keyword.  Use a function's argument list to pass in all parameters.

[ManiacDan]

Also, slight digression, but never use the 'global' keyword.  Use a function's argument list to pass in all parameters.

That too.

[Glese]

So at least two of those are unset.  See how that works?  Your array is randomly one or more of those 4 items, yet your function assumes all 4 will be there.

 

 

 

 

             if (isset($likes)) 
                    $array[] = $likes;
                if (isset($dislikes))
                    $array[] = $dislikes;
                if (isset($con_id))
                    $array[] = $con_id;
                if (isset($user_id)) {
                    $array[] = $user_id;
                }

 

 

What I do not understand is, the code is stating IF it is set, do assign it to the array, so if it is not set, do not assign it to the array, in this sense, I should not get the notice, as I stated I usually, do not get the notice only in a specific situation, when I do choose certain entries from a drop down menu, and with other entries I do not get the error.

[Glese]

I solved it by putting an isset condition check around everything possible.

[ManiacDan]

You're exactly right about the first part, then you get wrong.

 

When you create the array, you do an isset call and only add to the array when the item is set.  Yes, exactly right.

 

Then you attempt to use the fourth element in the array.  What happens if that's not set?  You get the error.

 

Also note that this code has no way of telling which variable is which in this array.  You assume that $array[2] is con_id.  What if $dislikes isn't set?  Then con_id will be in $array[1]

 

You need to rethink this whole thing and look into naming your array keys, so you have $array['con_id'] instead of $array[2]

[Glese]

 

 

You need to rethink this whole thing and look into naming your array keys, so you have $array['con_id'] instead of $array[2]

Is that what you would recommend in general, or only specifically in this case?

 

Thanks for the recommendation though, I did not even notice that one.

[ManiacDan]

In general, if you're storing specific variables in an array, name the indexes so you know which is which.  If you're storing a bunch of variables that are all the same kind, you don't need to name them.