MySQL PHP record checking - php

Question

im having a simple mysql/php problem. so i am adding in Image titles for my website, and the code is displayed below. It works, but when you dont put a image, it shows up as blank. I need it to show up as 'No image title' (bc i will use this for image description to). It basically gets the image name, then takes the title from that row.
So how do i do it? :/ im still very new to PHP.
<?php
if (isset($imgtitleset))
{
$sql = "SELECT image_title FROM images WHERE image_name = '$image_main'";
$result = mysql_query ($sql);
while ($row = mysql_fetch_array($result))
{
$imgtitle= $row["image_title"];
echo "$imgtitle";
}
}
else {
echo 'no image title';
}
?>

Answer 1

Change the while loop like so:
while ($row = mysql_fetch_array($result)) {
$imgtitle= $row["image_title"];
if($imgtitle != '') {
echo $imgtitle;
} else {
echo 'no image title';
}
}
Also, I'm not sure what the $imgtitleset variable is for, but you can probably get rid of the if statement checking to see whether it's set.
Edit: the whole thing should probably look like this:
<?php
$sql = "SELECT image_title FROM images WHERE image_name = '$image_main'";
$result = mysql_query ($sql);
while ($row = mysql_fetch_array($result)) {
$imgtitle= $row["image_title"];
if($imgtitle != '') {
echo $imgtitle;
} else {
echo 'no image title';
}
}
?>

Answer 2

This all depends on what $imgtitleset is equal to. It is clearly set against something:
while ($row = mysql_fetch_array($result)) {
$imgtitle = $row["image_title"];
if (isset($imgtitle))
echo "$imgtitle";
else
echo 'no image title';
}
This would mean if nothing was found in the database then it will echo the no image title. However like I said, this could depend on what $imgtitleset is, maybe post the code for that?

Answer 3

If you only expect the select to return a single row, then use if rather than while and return the error on else:
<?php
if (isset($imgtitleset))
{
$sql = "SELECT image_title FROM images WHERE image_name = '$image_main'";
$result = mysql_query ($sql);
if ($row = mysql_fetch_array($result))
{
$imgtitle= $row["image_title"];
echo "$imgtitle";
}
else {
echo 'no image title';
}
}
?>