Undefined Index PHP, MySQL, PhpMyAdmin - php

Question

Error:
( ! ) Notice: Undefined index: VIN in C:\wamp\www\php\myJoyofPHP\viewcar.php on line 10
09 <?php include 'db.php';
10 $vin = $_GET['VIN'];
11 $query = "SELECT * FROM INVENTORY";
12 /* Try to query the database */
13 if($result = $mysqli->query($query)){
14 //Don't do anything if successful.
15 }
16 else
17 {
18 echo "Sorry, a vehicle with VIN of $vin cannot be found ".mysql_error()."<br>";
19 }
20
21 //Loop through all the rows returned by the query, creating a table row for each
22 while ($result_ar = mysqli_fetch_assoc($result)){
23 $year = $result_ar['YEAR'];
24 $make = $result_ar['Make'];
25 $model = $result_ar['Model'];
26 $trim = $result_ar['TRIM'];
27 $color = $result_ar['EXT_COLOR'];
28 $interior = $result_ar['INT_COLOR'];
29 $mileage = $result_ar['MILEAGE'];
30 $transmission = $result_ar['TRANSMISSION'];
31 $price = $result_ar['ASKING_PRICE'];
32 }
33 echo "$year $make $model </p>";
34 echo "<p>Asking Price: $price </p>";
35 echo "<p>Exterior Color: $color </p>";
36 echo "<p>Interior Color: $interior </p>";
37
38 $mysqli->close();
39 ?>
I'm not concerned about security at the moment because I am just learning the basics as of now. So please don't respond saying that it's vulnerable code, I know.
I have a Column named 'VIN' in the 'cars' database within PhpMyAdmin with multiple values in it but For some reason this error comes up. Also, underneath the error, a single data set appears.
Why would I be getting this error?
Why would I be getting the error especially if a database exists with the same column name 'VIN'?
When I enter the following code above it, there is a variable error too.
$value = isset($array['VIN']) ? $array['VIN'] : '';
$value = array_key_exists('VIN', $array) ? $array['VIN'] : '';
Also, this is the db.php file that is included:
<?php
//localhost, root, password, database
$mysqli = new mysqli('localhost','root','','cars');
/*check connection */
if (mysqli_connect_errno()){
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
//select a database to work with
$mysqli->select_db("cars");
?>
As I mentioned, I understand this is a vulnerable database. There is no password. Please don't post anything regarding that, as I need to fully understand the basics. Thank you.

Answer 1

First, is not an "error", it is a "notice". You can ignore it almost all the time.
Its mean 'VIN' index is not defined inside the $_GET global var/array.
That is 'VIN' do not come in the url parameters.
PHP assumes it like an empty string.
Maybe the best solution is some like:
if( isset($_GET['VIN']) && $_GET['VIN'] != "" ){ // you can add others validations like string lenght or regex
$vin = $_GET['VIN'];
$query = "SELECT * FROM INVENTORY";
//...all your code...
}else{
// Do something when no VIN comes in the url
}
Obviously you need to use your script with de VIN parameter:
.....viewcar.php?VIN=ABC123

Answer 2

You can use the following code and notice message will disappear, its just a notice saying that $vin is not defined because sometimes $vin is empty, it does not get any value from $_GET
if (isset($_GET['VIN']))
{
$vin = $_GET['VIN'];
}

Answer 3

This line is causing you to receive that warning $vin = $_GET['VIN']; which means that you are not receiving VIN in the request, you need to check it as
if (isset($_GET['VIN'])) {
//rest of the code
}

Answer 4

Short and simple:
$vin = #$_GET['VIN'];
If VIN was not fed in
There will be no error message;
$vin will be NULL.
Later, you can use if (isset($vin)) ..., which reads easily.