php - display array only if exists -

i made page users can see data entered them in database.
used:

$select = "select * texts user='".$user."' order date desc, id desc"; $result = mysql_query($select); $array = array(); while($show = mysql_fetch_assoc($result)) { $array[] = $show; } echo "<strong>".$array[0]['id']."</strong><br />"; echo "<strong>".$array[1]['id']."</strong><br />"; echo "<strong>".$array[2]['id']."</strong><br />"; echo "<strong>".$array[3]['id']."</strong><br />"; echo "<strong>".$array[4]['id']."</strong><br />"; 

code works, have less 10 values return​​, more.
if use , have 2 arrays return, get:

notice: undefined offset: 2 in ownposts.php on line 15 notice: undefined offset: 3 in ownposts.php on line 16 notice: undefined offset: 4 in ownposts.php on line 17 

how it's possible echo $arrray[4]['id] if exist $array[4]?
i've tried with:

$zero = $array[0]; if(!empty($zero)) { echo "<strong>".$zero['id']."</strong><br />"; } $four = $array[4]; if(!empty($four)) { echo "<strong>".$five['id']."</strong><br />"; } 

but doesn't work excepted , still return notice: undefined offsed: 4 in ownposts.php on line 17.

instead of do, this:

while($show = mysql_fetch_assoc($result)) { $array[] = $show; } echo "<strong>".$array[0]['id']."</strong><br />"; echo "<strong>".$array[1]['id']."</strong><br />"; echo "<strong>".$array[2]['id']."</strong><br />"; echo "<strong>".$array[3]['id']."</strong><br />"; echo "<strong>".$array[4]['id']."</strong><br />"; 

why not show whatever found mysql find it:

while($show = mysql_fetch_assoc($result))   {     echo "<strong>".$show['id']."</strong><br />";   } 

or if have other things array code showed us, use loop on newly created array, ie foreach.