Show all items from a database even without a picture - php

Question

I'm making a webpage that displays my database. I'm using 2 tables for this.
Table 'artikel' and 'images'
Both tables got a key that link to each other: A_ARTCODE and I_ARTCODE
Both keys are the same.
Now I got the problem my keys were not linked and I saw this was, because I didn't used this:
WHERE artikel.A_ARTCODE = images.I_ARTCODE
So first my page looked like this: click here
and when I used that code I got this: click here
Like you see, now my both ID's are correct.
But I was talking about my tables, in the table 'images' I only got 2 images. Those you see now, but I got like 383 items in my 'artikel' table.
My question now is, how can I display them all? Even the items without a picture?
This is my total php code:
<?php
include('connect-mysql.php');
if (!empty($_GET["page"])) {
$page = $_GET["page"];
} else {
$page=1;
};
$start_from = ($page-1) * 20;
$sqlget = "SELECT *
FROM artikel, images
WHERE artikel.A_ARTCODE = images.I_ARTCODE
LIMIT $start_from, 20
";
$sqldata = mysqli_query($dbcon, $sqlget) or die('error getting');
define('IMAGE_PATH', 'images/');
$sql = "SELECT COUNT(A_ARTCODE) FROM artikel";
$rs_result = mysqli_query($dbcon, $sql) or die ("mysqli query dies");
$row = mysqli_fetch_row($rs_result) or die ("mysqli fetch row dies");
$total_records = $row[0];
$total_pages = ceil($total_records / 20);
echo "<div class='boven'>";
for ($i=1; $i<=$total_pages; $i++) {
echo "<a href='index.php?page=".$i."'>".$i."</a> ";
};
echo "</div>";
echo "<table>";
echo "<tr><th>A_ARTCODE</th><th>A_NUMMER</th><th>A_OMSCHRN</th><th>A_REFLEV</th><th>A_WINKEL</th><th>I_ARTCODE</th><th>I_FILE</th></tr>";
while($row = mysqli_fetch_array($sqldata)){
echo "<tr><td align='right'>";
echo $row['A_ARTCODE'];
echo "</td><td align='left'>";
echo $row['A_NUMMER'];
echo "</td><td align='left'>";
echo $row['A_OMSCHRN'];
echo "</td><td align='left'>";
echo $row['A_REFLEV'];
echo "</td><td align='right'>";
echo $row['A_WINKEL'];
echo "</td><td align='right'>";
echo $row['I_ARTCODE'];
echo "</td><td align='right'>";
echo "<img src='". IMAGE_PATH . $row['I_PATH']. '/' . $row['I_ID']. '.png' ."' />";
echo "</td></tr>";
}
echo "</table>";
echo "<div class='onder'>";
for ($i=1; $i<=$total_pages; $i++) {
echo "<a href='index.php?page=".$i."'>".$i."</a> ";
};
echo "</div>";
?>

Answer 1

To achieve this U could use a LEFT JOIN sql.
This will give u all records for the articles, even when no images are defined for it
SELECT *
FROM artikel a
LEFT JOIN images i
ON a.A_ARTCODE = i.I_ARTCODE
LIMIT $start_from, 20

Answer 2

One way is to use a left join.
select *
from artikel
left join images
on artikel.A_ARTCODE = images.I_ARTCODE
LIMIT $start_from, 20;
Break yourself of the habit of writing things like from artikel, images; this isn't 1989. Use ANSI joins.
Another way is to guarantee that every row has an image. Use a default image (and a default value with a "not null" constraint for it in the table). It might be just a gradient with the text "Picture coming soon" on it.