IF ISSET not working?

IN DB

row1 (type column=apple), (stage column=1)
row2 (type column=orange), (stage column=NULL)

IN PHP

$query = mysql_query("SELECT * from fruits");

$row = mysql_fetch_array($query);

    $num = mysql_num_rows($query);
    $i=0;
    $storeMyData = array();
    while($row = mysql_fetch_array($query))
    {
        if(isset($row['type'])){
            $type = "-" . $row['type'];
        } else {
            $type = NULL;
        }

        if(isset($row['stage'])){
            $stage = "STAGE " . $row['stage'];
        } else {
            $stage = NULL;
        }
        $fruit = implode (", ", array_filter(array($type, $stage)));

    // This will show 2 rows of data
    $storeMyData[] = $fruit;  

    // store current data in array
    $i++;
    }

    /* this will echo your storedData of loop */
    foreach($storeMyData as $prevData)

    /* or join the data using string concatenation */
    $allFinalData2 = "";

    /* this will echo your storedData of loop */
    foreach($storeMyData as $prevData)
    {
        $allFinalData2 = $allFinalData2.$prevData ;  // keep on concatenating
    }

    echo $allFinalData2;

The final output shows as "Apple, Stage 1" "Orange, Stage".

How can I show it "Apple, Stage 1" "Orange" without the word Stage if stage is NULL for row2 (orange) in DB.

Change this line

while($i < $num)

to this

while($row = mysql_fetch_array($query))

also, there is no need to call mysql_query inside your loop, use this to prevent confusion.

$type = "-" . $row['type'];
$stage = "STAGE " . $row['stage'];