while mysql_fetch_array loop doesn't run - php

Question

Here is my code
$itemSelect="select * from items where gender='men' and type='suite'";
$itemQuery=mysql_query($itemSelect) or die(mysql_error());
$num=count($itemQuery);
echo $num;
if($itemQuery)
{
echo "I am here";
echo "<table>";
while($row = mysql_fetch_array($itemQuery))
{
$photo=$row['photo'];
$name=$row['name'];
$price=$row['price'];
echo "<tr><td>". $photo ."</td><td>" .$name ."</td><td>" .$price ."</td> </tr>";
}
echo "</table>";
}
else
{
echo "There is some mistacke";
die();
}
So here $num=count($itemQuery); shows me that there is 1 item that satisfies the search but the loop is never executed what could be the problem? Thanks in advance.

Answer 1

The reason you are getting 1 is because of this code:
$num=count($itemQuery);
echo $num;
count is a function to get the number of items in an array. If you pass anything that isn't an array (or doesn't implement the ICountable interface), then count returns always 1.
So this piece of code is incorrect in that it shows 1, while that number isn't at all related to the number of rows returned.
The query itself looks fine to me, although I cannot validate it against your actual database. If you see the message "I am here", then your query executed fine, but didn't return any rows. If you don't see that message, then something else went wrong. You can use mysql_error to try to find any errors in that case.

Answer 2

$count= mysql_num_rows($itemQuery);
echo $count;
if($count > 0)){
echo "<table>";
while($row = mysql_fetch_array($itemQuery)){
$photo=$row['photo'];
$name=$row['name'];
$price=$row['price'];
echo "<tr><td>". $photo ."</td><td>" .$name ."</td><td>" .$price ."</td> </tr>";
}
echo "</table>";
}else{
echo "there are no results";
die();
}