Undefined Index PHP - php

Question

I've done some searching but can't seem to work out the answer. I'm new to PHP.
Here is my HTML form;
<form id = "getGame" action="Form.php" method = "get">
<label> Serial
<input type = "search" name = "serial" />
</label>
<label> Title
<input type = "search" name = "title" />
</label>
<label> Year
<select name="gameYear">
<option value="1998">1998</option>
<option value="1999">1999</option>
<option value="2000">2000</option>
</select>
</label>
<label> Price
<input type = "search" name = "price" />
</label>
<input type = "submit" name="search" value = "Search">
</form>
Here is my PHP page;
include 'db_Conn.php';
$gameYear = $_GET ['gameYear'];
$sql = "SELECT serialNo, gameTitle, gameYear gamePrice FROM games WHERE gameYear = '$gameYear'";
$Games = mysqli_query($conn, $sql)
or die(mysqli_error($conn));
while ($row = mysqli_fetch_assoc($Games)) {
$serial = $row['serialNo'];
$title = $row['gameTitle'];
$gameYear = $row['gameYear'];
$gamePrice = $row['gamePrice'];
echo "<div>$serial, $title, $gameYear, $gamePrice</div>\n";
}
mysqli_free_result($Books);
mysqli_close($conn);
?>
My problem is when I run the form I get the following message "Notice: Undefined index: gameYear". However, on the line below it displays all the records from the game year I selected. Example of what displays below;
Notice: Undefined index: gameYear in ____ on line 40
493928, Test Drive, 2002, $12.95
Thanks, any point in the right direction is appreciated.

Answer 1

You missed one ,. Change it to:
$sql = "SELECT serialNo, gameTitle, gameYear, gamePrice FROM games WHERE gameYear = '$gameYear'";

Answer 2

if no data found in the database it will produce 4 errors . so first of all check any data found or not by mysqli_num_rows() function.
if (mysqli_num_rows($Games) > 0) {
while ($row = mysqli_fetch_assoc($Games)) {
$serial = $row['serialNo'];
$title = $row['gameTitle'];
$gameYear = $row['gameYear'];
$gamePrice = $row['gamePrice'];
echo "<div>$serial, $title, $gameYear, $gamePrice</div>\n";
}
}
Yes of course you missed one comma so sql should be like this
$sql = "SELECT serialNo, gameTitle, gameYear, gamePrice FROM games WHERE gameYear = '$gameYear'";