Passing specific of multiple sql query output to form dynamically - php

Question

The query works fine. I get a table with the associated data in rows. As you see I'd like a button with each row that calls a function that affects that specific row with which it is echoed. I have it working with a normal form, that is, If I put the value of Part_Number in manually the function will run successfully, but the issue is to have it more dynamic. I provided the applicable code.
In short: When I push the button that is echoed with the Part_Number it should update (add one) to the Quantity of that Part_Number.
The SQL doesn't throw an error but it also doesn't work. Everything except passing the variables dynamically works.
$sql = "SELECT id, Part_Number, Description, Location, Quantity, Certificate
FROM Inventory WHERE Part_Number LIKE'%$q%'";
$result = $conn->query($sql);
while($row = $result->fetch_assoc())
echo "
<form method='POST' action='plusone.php'><tbody>
<tr>
<th><h2>".$row["Part_Number"]."</h2></th>
<td><h2>".$row["Description"]."</h2> </td>
<td><h2>".$row["Location"]."</h2> </td>
<td>
<input type='submit' class='btn btn-success' value='Plus 1'>
<input type='hidden' class='form-control' name='Part_Number_Entry'
id='$Part_Number_Entry' value=''>
</tr>
</tbody>
</form>";
if (isset($Part_Number1)) {
// the single quotes around the last variable are very important. If left out
the update wont work
$sql = "UPDATE TestTable SET Quantity= Quantity + 1 WHERE Part_Number=
'$Part_Number1'";
}elseif (!isset($PartNumber)) {
echo "PartNumber is not set";
# code...
}
if (mysqli_query($conn, $sql)) {
echo "Added One to current Record successfully";
} else {
echo "Error updating record: " . mysqli_error($conn);
echo $Quantity;
echo " ";
}
mysqli_close($conn);
?>

Answer 1

there are quite a few issues but this is the least code i could come up wit that will hopefully get you at least going in the right direction:
<input type='hidden' class='form-control' name='Part_Number_Entry' id='$Part_Number_Entry' value=''>
should be
<input type='hidden' class='form-control' name='Part_Number_Entry' value='$row["Part_Number_Entry"]' >
its the 'name' that becomes the array key and the value is from your db row
if (isset($Part_Number1)) {
should be
if (isset($_POST['Part_Number_Entry')) {
your posting the form, and the value you want is in that array.
To get $Part_Number1 for the query
$Part_Number1=(int)$_POST['Part_Number_Entry');
cast as int by you need to much safer db queries in the future.
This is not a great answer and i hope some posts something better.