Select all data after selecting another option - php

Question

I am trying to create a submission on a form where I can show all results from a table as well as show individual results. I can achieve the page load to show all until the form is submitted, however when I then try and select all again im struggling.
On page load im simply doing:
<?php
if (isset($_POST['submit'])) {
$teamData = $_POST['teamData'];
var_dump($teamData);
$sql = "SELECT * FROM team WHERE dashboardId = 1 AND id = $teamData";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
$teamName = $row['name'];
}
}
} else {
$teamName = 'All';
echo 'no submission yet';
}
?>
Setting the variable to say 'all'
<p>Team: <?php echo $teamName; ?></p>
Once an option has been selected it check the database and uses the name of and sets it. However in the dropdown list if i want to show all results again i get an error of:
Undefined variable: teamName
which makes sense because of my form:
<form method="POST" action="">
<select name="teamData">
<option selected value="" disabled>Select your team</option>
<option value="allTeamData">All team data</option>
<?php teamMembers(); ?>
</select>
<input type="submit" name="submit" value="Submit" />
</form>
I am just struggling to understand the logic of how to select all again from the drop down.

Answer 1

$teamName = 'All'; // Initialise the variable at top
// Change the condition to this for better restriction
if ($_POST && isset($_POST['submit'])) {
$teamData = $_POST['teamData'];
var_dump($teamData);
$sql = "SELECT * FROM team WHERE dashboardId = 1 AND id = $teamData";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
$teamName = $row['name'];
}
}
}