PHP & SQL - Update query fails to update value - php

Question

I'm trying to create a very easy stock managing system. I'm able to show all the items in my table 'parts' and i'm showing the amount in a textbox. However, when i change the value from, for example, 0 to 5 in the textbox and i press my submit button, it doesn't update the stock.
Below is my code, i don't have alot of experience with update querys but i've read about it on php.net, obviously.
<?php
echo "<table width=\"800\" class=\"nieuws\">";
$db=mysqli_connect("localhost","root","","lichtwinkel");
$p=mysqli_query($db, "SELECT * FROM parts WHERE product LIKE 1");
echo "<form method='post' action=''>";
echo "<tr><th></th><th>Onderdeel nummer</th><th>Eigenschappen</th><th>Prijs</th><th>Voorraad</th></tr>";
while ($row = mysqli_fetch_array($p)){
echo "<tr>";
echo "<td><img class='lamp' src='../css/images/".trim($row['partnr']).".png' alt='Geen afbeelding beschikbaar'></td>";
echo "<td>".$row['partnr']."</td>";
echo "<td>".$row['specs']."</td>";
echo "<td>€ ".$row['price']."</td>";
echo "<td><input type='text' id='aantal' name='aantal' value=$row[voorraad] /></td>";
echo "<td><input type='submit' id='update' name='update' value='Update' /></td>";
echo "</tr>";
}
echo "</table>";
if(isset($_POST['aantal']) && $_POST['update']) {
$y = $_POST['aantal'];
$p=mysqli_query($db, "UPDATE parts SET voorraad = '$y' WHERE partnr = $row[0]");
}
echo "</form>"
?>
Simply said, what i'm trying to achieve is the following:
Whenever i change the value displayed in the texbox, and i press my submit button, i want it to update the value in the database.
Does anyone know what i'm doing wrong? Any ideas? Articles i should read?
All help would be appreciated.
Thank you.

Answer 1

As i see, you were doing it wrong at all.
First you can't use form tag within more then one td element.
You were didn't close the form tag, only at end. (So if it loops 6 times, you will have 6 forms open, but one ended!).
At update, you're selecting row[0] - it's outside of loop with rows?
Even if you update it, it will show wrong results again. Update should be above selects! So it picks up newly updated value.
What to do:
First make one form for all updates.
Use your submit button to have value DATABASE_ID.
Make the name of "aantal" to "aantalDATABASE_ID".
At submit check for $_POST['update'], and use it's value (DATABASE_ID) to get input $_POST["aantal".$_POST['update']].
Do update, you have all you need.
Example:
<?php
echo "<form method='post' action=''>";
echo "<table width=\"800\" class=\"nieuws\">"
$db=mysqli_connect("localhost","root","","lichtwinkel");
if(isset($_POST['update']) && !empty($_POST['update'])) {
$y = $_POST['aantal'.$_POST['update']];
$p=mysqli_query($db, "UPDATE parts SET voorraad = '".$y."' WHERE partnr = '".$_POST['update']."'");
}
$p=mysqli_query($db, "SELECT * FROM parts WHERE product LIKE 1");
echo "<tr><th></th><th>Onderdeel nummer</th><th>Eigenschappen</th><th>Prijs</th><th>Voorraad</th></tr>";
while ($row = mysqli_fetch_array($p)){
echo "<tr>";
echo "<td><img class='lamp' src='../css/images/".trim($row['partnr']).".png' alt='Geen afbeelding beschikbaar'></td>";
echo "<td>".$row['partnr']."</td>";
echo "<td>".$row['specs']."</td>";
echo "<td>€ ".$row['price']."</td>";
echo "<td><input type='text' id='aantal' name='aantal".$row[0]."' value='".$row[voorraad]."' /></td>";
echo "<td><input type='submit' id='update' name='update' value='".$row[0]."' /></td>";
echo "</tr>";
}
echo "</table>";
echo '</form>';
?>
After all, take care about SQL Injections. "aantal" value is user input. As the submit value can be changed.