PHP MySQL UPDATE query not posting value from text box - php

Question

I have a form with a select list populated from a database. When the query is executed, a success message is returned however the database is not updated.
Here is the form:
<form id="formPrice">
<?php
$conn = new mysqli('localhost', 'something', 'something', 'something')
or die ('Cannot connect to db');
$result = $conn->query("SELECT PlotNumber FROM Developments WHERE Development = 'GREENGRAVES' AND Price = 'BOOKED'");
echo "<select name='plot_update'>";
while ($row = $result->fetch_assoc()) {
echo "<option value=\"Plot\">" . $row['PlotNumber'] . "
</option>";
}
echo "</select>";
?>
<input name="price" type="text" id="price">
<input name="update" type="submit" class="update_price">
</form>
Here is the ajax request:
$('.update_price').click(function() {
var FormData = $('form').serialize();
$.ajax({
type: "POST",
url: '../php/update_price.php',
data : FormData,
success:function(html){
document.getElementById("result_two").innerHTML=html;
}
});
return false;
});
Here is the php
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$price = $_POST['price'];
$plot = $_POST['plot'];
$sql = "UPDATE Developments SET Price = '".$price."' WHERE PlotNumber = '".$plot."' ";
if ($conn->query($sql) === TRUE) {
echo "Record updated successfully";
} else {
echo "Error updating record: " . $conn->error;
}
$conn->close();
The page displays "record updated successfully" in the result div, however database entry remains unchanged. Any ideas why? I'm stumped.

Answer 1

The error seemed to be with this line in the front end form:
echo "<option value=\"Plot\">" . $row['PlotNumber'] . "</option>";
When I changed to:
echo "<option>" . $row['PlotNumber'] . "</option>";
it worked!

Answer 2

There is issues with form field under php script:
change your php script to the following:
<form id="formPrice">
<?php
$conn = new mysqli('localhost', 'something', 'something', 'something')
or die ('Cannot connect to db');
$result = $conn->query("SELECT PlotNumber FROM Developments WHERE Development = 'GREENGRAVES' AND Price = 'BOOKED'");
echo "<select name='plot'>";
while ($row = $result->fetch_assoc()) {
echo "<option value=" . $row['PlotNumber'] . ">" . $row['PlotNumber'] . "</option>";
}
echo "</select>";
?>
<input name="price" type="text" id="price">
<input name="update" type="button" class="update_price">
Always make sure - Select option value needs to have the actual value of the record and select field will have the form field name.