How do i define from which table that specific column comes from? - php

Question

I made a page in which one can edit the prices for each article that each webshop has. The price that it already has is in the textfield and it can be changed to whatever the user may want. Then the user hits the Edit (Wijzigen is Dutch for Edit) button and it should change to whatever was in the textfield before pressing it.
But I'm having quite some trouble with the UPDATE query since I need to give it some ID's to select the right one. And, as you'll be able to see in the source, right after starting the while() loop, I can't specify the right ID from the right table.
I would appreciate it if you could help me out with what I need to change, add or delete to make this work.
<?php
include 'inc/inc.php';
$artikel = mysqli_query($mysql, "SELECT a.Artikel_ID, a.Productnaam, aw.Artikel_ID, aw.Webshop_ID, aw.Prijs, w.Webshop_ID, w.Shopnaam FROM artikel a, artikel_webshop aw, webshops w WHERE a.Artikel_ID = aw.Artikel_ID AND aw.Webshop_ID = w.Webshop_ID GROUP BY aw.Webshop_ID, aw.Artikel_ID");
echo '<table id="specialtable"><tr style="background-color:#F8F8F8;"><td>Productnaam</td><td>Webshop</td><td>Prijs</td><td>Wijzigen</td></tr>';
$i = 0;
$j = 1;
while($artikelR = mysqli_fetch_assoc($artikel)) {
$artikel_id = $artikelR['Artikel_ID'];
$webshop_id = $artikelR['Webshop_ID'];
echo '<form method="post">';
if($i == 0){
echo '<tr style="background-color:#DDD;">';
$i = $i + 1;
}else{
echo '<tr style="background-color:#F8F8F8;">';
$i = $i - 1;
}
echo '<td>' . $artikelR['Productnaam']. '</td><td>' . $artikelR['Shopnaam'] . '</td><td>€ <input type="text" name="' . $j . '" value="' . $artikelR['Prijs']. '"></td>';
echo '<td><input type="submit" name="' . $j . '" value="Wijzigen"></td></tr></form>';
$jNew = $_POST['' . $j . ''];
$j++;
$test = $_POST['' . $j . ''];
if(isset($test)) {
$sql = mysqli_query($mysql, "
UPDATE artikel_webshop
SET Prijs = '" . $jNew . "'
WHERE Artikel_ID = '$artikel_id'
AND Webshop_ID = '$webshop_id'");
echo 'U heeft succesvol de prijs van het artikel ' . $artikelR['Productnaam'] . ' van de webshop ' . $artikelR['Shopnaam'] . ' gewijzigd.<br><br>';
$url = 'prijsWij.php';
echo '<meta http-equiv="refresh" content="4;URL=' . $url . '">';
}
}
echo '</table></form>';
?>
Edit
I finally got it to work. I'm pretty bad at explaining, so I'll just add the new code:
<?php
include 'inc/inc.php';
$artikel = mysqli_query($mysql, "
SELECT a.Artikel_ID, a.Productnaam, aw.Artikel_ID as awa_id, aw.Webshop_ID as aww_id, aw.Prijs, w.Webshop_ID, w.Shopnaam
FROM artikel a, artikel_webshop aw, webshops w
WHERE a.Artikel_ID = aw.Artikel_ID
AND aw.Webshop_ID = w.Webshop_ID");
echo '<table id="specialtable"><tr style="background-color:#F8F8F8;"><td>Productnaam</td><td>Webshop</td><td>Prijs</td><td>Wijzigen</td></tr>';
$i = 0;
$j = 1;
while($artikelR = mysqli_fetch_assoc($artikel)){
$artikel_id = $artikelR['awa_id'];
$webshop_id = $artikelR['aww_id'];
echo '<form method="post">';
if($i == 0){
echo '<tr style="background-color:#DDD;">';
$i = $i + 1;
}else{
echo '<tr style="background-color:#F8F8F8;">';
$i = $i - 1;
}
echo '<td>' . $artikelR['Productnaam']. '</td><td>' . $artikelR['Shopnaam'] . '</td>';
echo '<td>€ <input type="text" name="prijsTest" value="' . $artikelR['Prijs']. '"></td>';
echo '<td><input type="submit" name="' . $j . '" value="Wijzigen"></td></tr></form>';
$kNew = $_POST['prijsTest'];
$test = $_POST['' . $j . ''];
$j = $j + 1;
if(isset($test)){
$sql = mysqli_query($mysql, "
UPDATE artikel_webshop
SET Prijs = " . $kNew . "
WHERE Artikel_ID = '$artikel_id'
AND Webshop_ID = '$webshop_id'");
echo 'U heeft succesvol de prijs van het artikel ' . $artikelR['Productnaam'] . ' van de webshop ' . $artikelR['Shopnaam'] . ' gewijzigd.<br><br>';
$url = 'prijsWij.php';
echo '<meta http-equiv="refresh" content="4;URL=' . $url . '">';
}
}
echo '</table></form>';
?>

Answer 1

SELECT a.Artikel_ID as a_id, a.Productnaam, aw.Artikel_ID as aw_id ...

Answer 2

In order to know which article was updated, you need to keep track of it in your form so that when the form gets submitted, you will be able to update the correct article.
It seems like you have one form per article. In this case, it's easy. When you generate your form, add a hidden input containing the article ID:
<input type="hidden" name="article_id" value="<?php echo $articleId; ?>"/>
When the form will be submitted, the article ID will be posted along with the new price. You can retrieve it through $_POST['article_id']. Then, you can use this variable in your MySQL query to update the right article!