Notice: Undefined index:

skallsen

Here's what I know:

I know that these errors only show up on my localhost pages because I have all the error reporting on. But I want to code this correctly anyway.

I've seen the posts about using isset and I understand basically how to use it, but I'm having trouble applying it to my code when using three possible variables to the page.

At the top of my page (products.php?variable=number) I have:

<?php

            $id=$_GET['id'];
	$productid=$_GET['productid'];
	$mfgid=$_GET['mfgid'];

for each possible type of id I have separate instructions farther down in the code:

if (isset($id)){
$queryid = "SELECT * FROM products2 WHERE products2ID=$id";
$resultid = mysql_query($queryid) or die(mysql_error());

}elseif (isset($productid)){

			$queryprodid = "SELECT * FROM products2 WHERE products2ID=$productid";

}elseif (isset($mfgid)){

			$querymfg = "SELECT * FROM products2 WHERE products2MfgID=$mfgid";

And at the bottom:

}else{
echo "not working";
}

I have one other undefined index in the center of my code that I don't know how to fix:

// Query table and get video file names. //
$sql = "SELECT products2Name,products2Movie FROM products2 WHERE products2ModelID=$subcat";

I'd really appreciate it if someone could tell me what I need to do to accurately test all my variables! Thanks!

dalecosp

if (isset($_GET['id'])) {
   $id=$_GET['id'];
}

As for the last one, it needs to be initialized before use, doesn't it?

bradgrafelman

@: If you do that, you don't really mitigate the issue at all - instead of using undefined indexes with $_GET['id'], you'd be using an undefined variable $id.

Instead, I like to solve this using the ternary operator (for compactness):

$foo_num = (isset($_COOKIE['foo_num']) ? (int)$_COOKIE['foo_num'] : NULL);

Since that not only a) lets me do some generic type of sanitizing/formatting to provided values as well as provide a default value when the value isn't provided.