How to get ID on a specific row in PHP and MySQL - php

Question

PHP is giving me error of
Notice: Undefined index: product_id in C:\xampp\htdocs\odieinventory\admin\add_sales.php on line 6
This is my query
<?php
if (isset($_POST['addCart'])) {
$product_id = $_POST['product_id'];
$qtyBuy = $_POST['qtyBuy'];
$addQuery = "INSERT INTO sales (product_id, quantity) VALUES ($product_id, $qtyBuy)";
$execQuery = mysqli_query($connection, $addQuery);
}
?>
This is my table
<form action="add_sales.php" method="POST">
<?php
$query = "SELECT * FROM products";
$exec = mysqli_query($connection, $query);
while ($row = mysqli_fetch_array($exec)) {
$product_id = $row['product_id'];
$product_name = $row['product_name'];
$description = $row['description'];
$product_quantity = $row['quantity'];
$product_price = $row['sell_price'];
?>
<tr>
<td class="text-center"><?php echo $product_id; ?>
<input type="hidden" name="product_id" value="<?php echo $product_id; ?>">
</td>
<td><?php echo $product_name; ?></td>
<td><?php echo $description; ?></td>
<td><?php echo $product_price; ?></td>
<td><?php echo $product_quantity; ?></td>
<td><input type="number" min="1" max="999" name="qtyBuy"></td>
</tr>
<?php } ?>
</tbody>
</table>
</div>
<div class="form-group">
<input type="submit" name="addCart" value="Add Items to Cart" class="btn btn-info pull-right">
</div>
</div>
</form>
I would also like to ask for advice on how to insert records on my cart. By clicking 'Add Items to Cart', I like to pass ONLY the items with values on the quantitiy input on the cart. How to achieve this with PHP?

Answer 1

In your table file table.php , you have to:
1- open the table and tbody HTML tags
2- make your html attributes form name as array (you have more than one input with the same name), in your case you are only submiting the last product_id and the last qtyBuy input values.
table.php
<form action="add_sales.php" method="POST">
<div><table><tbody>
<?php
$query = "SELECT * FROM products";
$exec = mysqli_query($connection, $query);
while ($row = mysqli_fetch_array($exec)) {
$product_id = $row['product_id'];
$product_name = $row['product_name'];
$description = $row['description'];
$product_quantity = $row['quantity'];
$product_price = $row['sell_price'];
?>
<tr>
<td class="text-center"><?php echo $product_id; ?>
<input type="hidden" name="product_id[]" value="<?php echo $product_id; ?>">
</td>
<td><?php echo $product_name; ?></td>
<td><?php echo $description; ?></td>
<td><?php echo $product_price; ?></td>
<td><?php echo $product_quantity; ?></td>
<td><input type="number" min="1" max="999" name="qtyBuy[]"></td>
</tr>
<?php } ?>
</tbody>
</table>
</div>
<div class="form-group">
<input type="submit" name="addCart" value="Add Items to Cart" class="btn btn-info pull-right">
</div>
</form>
In your add_sales.php file, you have to loop the $_POST['qtyBuy'] variable and check if the given quantity of each product is > 0
add_sales.php
<?php
if (isset($_POST['addCart']) && $_POST['addCart']=="Add Items to Cart") {
foreach($_POST['qtyBuy'] as $index=>$value){
if($value > 0){
$addQuery = "INSERT INTO sales (product_id, quantity) VALUES (".$_POST['product_id'][$index].", ".$value.")";
$execQuery = mysqli_query($connection, $addQuery);
}
}
}
?>

Answer 2

First of all, you should never put users input ($_POST) directly into the SQL to avoid SQL injection - use prepared statements with mysqli.
It would be easiest for you to add a separate form with product ID and quantity field and an "Add to cart" button for each product in the table.