How can I update data for a selected row? - php

Question

I'm learning PHP, html, and sql. I am trying to create a form that allows users to update data on existing customers. I was wondering what went wrong with my code because every time I tried to make changes on an existing customer, the changes are reflected on the first customer of the data set. For example, when I changed the data for customer ID #14, the changes are shown in customer ID #1. Customer ID #14 still holds the old data. Here is my code. Can someone give me some tips on how to fix it? I spent hours on this, but I don't know what I'm missing. Thanks a lot!
This is part of the code in the main.php file.
<div>
<form method="post" action="updatecustomer.php">
<fieldset>
<legend>Update Existing Customer</legend>
<li>First Name: <input type="text" name="fName"> Last Name: <input type="text" name="lName"</li>
<li>Email Address: <input type="text" name="email"></li>
<li>Phone Number: <input type="text" name="phone_number"></li>
<li>Street Number: <input type="text" name="address_no"> Street Line 1: <input type="text" name="address_street1"></li>
<li>Street Line 2 (Apt or Unit Number): <input type="text" name="address_street2"></li>
<li>City: <input type="text" name="address_city"> State: <input type="text" name="address_state"> Zip: <input type="text" name="address_zip"> </li>
<li>Customer ID:
<select name="customer_id">
<?php
if(!($stmt = $mysqli->prepare("SELECT customer_id, customer_id FROM customer"))){
echo "Prepare failed: " . $stmt->errno . " " . $stmt->error;
}
if(!$stmt->execute()){
echo "Execute failed: " . $mysqli->connect_errno . " " . $mysqli->connect_error;
}
if(!$stmt->bind_result($customer_id, $customer_id)){
echo "Bind failed: " . $mysqli->connect_errno . " " . $mysqli->connect_error;
}
while($stmt->fetch()){
echo '<option value=" '. $customer_id . ' "> ' . $customer_id . '</option>\n';
}
$stmt->close();
?>
</select>
</li>
</fieldset>
<input type="submit" name="update" value="Update Customer">
</div>
Here is the updatecustomer.php code:
if(!($stmt = $mysqli->prepare("UPDATE customer SET fName=?, lName=?, email=?, phone_number=?, address_no=?, address_street1=?,
address_street2=?, address_city=?, address_state=?, address_zip=? WHERE customer_id =?"))){
echo "Prepare failed: " . $stmt->errno . " " . $stmt->error;
}
if(!($stmt->bind_param("sssiissssii",$_POST['fName'],$_POST['lName'],$_POST['email'],$_POST['phone_number'], $_POST['address_no'],
$_POST['address_street1'],$_POST['address_street2'],$_POST['address_city'],$_POST['address_state'], $_POST['address_zip'], $_POST['customer_id']))){
echo "Bind failed: " . $stmt->errno . " " . $stmt->error;
}
if(!$stmt->execute()){
echo "Execute failed: " . $stmt->errno . " " . $stmt->error;
} else {
echo "Updated " . $stmt->affected_rows . " rows to customer.";
}

Answer 1

Not sure, but try to remove spaces here first value=" '. $customer_id . ' " because this will send you a string type value like string(4) " 14 ". I'd go for intval($_POST['customer_id']) next.