MySQL select where column LIKE and is not empty - php

Question

I have the following query and I'm trying to select only the rows where I have picture.
In table kat_gl_picture I have 3 categories, but I don't have picture in all 3 categories yet!
All work just fine, but I have printed name of third category, where I don't have picture.
I tried WHERE link LIKE '%$first_var%'AND NOT (link <=> NULL)
....IS NOT NULL - but nothing yet worked.
Tabele1 and 2 and web problem solved
<?
include("connection.php");
$kategorije = mysql_query("SELECT * FROM kat_gl_picture ORDER BY rbr");
while ($red=mysql_fetch_array($kategorije))
{
$first_var = $red['kat'];
$result = mysql_query("SELECT id, naziv, ime, tekst, username, link, file_name, datum FROM Tab_Pic_Pic
WHERE link LIKE '%$first_var%'
ORDER BY id");
echo '<table>';
echo '<tbody>';
echo $first_var;
echo '<tr>';
echo '<TD valign="top">';
while ($row=mysql_fetch_array($result))
{
list($x, $y) = getimagesize("admin /upload/".$row['file_name']);
if ($x>$y) {
$y=($y/$x)*150;
$x=150;
}
else
{
$x=($x/$y)*115;
$y=115;
}
$ID_broj = $row["id"];
$tekst_broj = $row["tekst"];
?>
<? echo '<img src="admin /upload/'.$row['file_name'].'" height="'.$y.'" width="'.$x.'"/>';?>
<?
}
echo '</td>';
echo '</tr>';
echo '</tbody>';
echo '</table>';
}

Answer 1

The problem is that I cannot determine if there is a foreign key in your Tab_Pic_Pic table. You can however join on a LIKE.
Try this:
SELECT * FROM Tab_Pic_Pic JOIN kat_gl_picture ON Tab_Pic_Pic.link LIKE CONCAT('%', kat_gl_picture.kat, '%') ORDER BY kat_gl_picture.kat, Tab_Pic_Pic, rbr
It should give you a list of all pictures including the kat_gl_picture.kat field which you can just use a local variable to detect change.
Would be easier to provide a more accurate example if I had the full table info. Standard method is to have a foreign key in Tab_Pic_Pic that references the corresponding primary key from the kat_gl_picture table.
Alternatively, if you simply want to omit the blank category it can be done in your PHP code like this:
if(mysqli_num_rows($result)>0){
echo '<table>';
echo '<tbody>';
echo $first_var;
echo '<tr>';
echo '<TD valign="top">';
while ($row=mysql_fetch_array($result))
{
list($x, $y) = getimagesize("admin /upload/".$row['file_name']);
if ($x>$y) {
$y=($y/$x)*150;
$x=150;
}
else
{
$x=($x/$y)*115;
$y=115;
}
$ID_broj = $row["id"];
$tekst_broj = $row["tekst"];
?>
<? echo '<img src="admin /upload/'.$row['file_name'].'" height="'.$y.'" width="'.$x.'"/>';?>
<?
}
echo '</td>';
echo '</tr>';
echo '</tbody>';
echo '</table>';
}
Wrapping the output of your HTML in a simple condition like this should illustrate that the script itself is inefficient. Trust me when I say it can all be accomplished with one query. Do not get frustrated. You'll figure it out. Programming is hard.