PHP MySQL generating unique random number - php

Question

I don't understand why my code isn't working. The connection works and everything else however when I try to generate a unique random number and check from the MySQL if the number is there it still prints out a random number but it's not UNIQUE. Could anyone help me thx?
Here's my code:
$num = rand(1,5);
$sel_query = "SELECT * FROM test";
$result2 = $con->query($sel_query);
$i = 1;
for (;$i<2; $i++)
{
while($row = mysqli_fetch_array($result2))
{
if ($row['id'] == $num)
{
$num = rand(1,5);
$i = 0;
}
}
}

Answer 1

This should work:
$is_unique = false;
$num = false;
while (!$is_unique){
$num = rand(1,5);
$sel_query = "SELECT id from test where id = " . $num;
$result2 = $con->query($sel_query) or die($conn->error);
if (!mysqli_fetch_array($result2)){
$is_unique = true;
}
}
echo "Unique number is " . $num;
But if there aren't any more possible unique numbers, it will loop forever.

Answer 2

I know this is a bit old, but I found this question after needing a similar answer. I've taken Jodes's answer and updated it slightly, so that it won't run forever, is a function that returns the number, and accepts a mysqli connection as $mysqli:
function getUniqueNumber($mysqli)
{
$is_unique = false;
$num = false;
$times_run = 0;
while (!$is_unique)
{
if($times_run > 10)
{
echo "Run too many times, dying.";
die();
}
$num = rand(1,5);
$sel_query = "SELECT id from test where id = " . $num;
$result2 = $mysqli->query($sel_query) or die($mysqli->error);
if (!mysqli_fetch_array($result2))
{
$is_unique = true;
}
$times_run++;
}
return $num;
}