how to obtain table check on (if) condition basis - php

Question

I have an attendance system where users could go in and out the campus, I finished it with this condition:
if($i == 0){
$in = '✔';
$out = '';
}
if($i > 0){
if (strtotime(substr($arr[$i][0],0,10)) == strtotime(substr($arr[$i-1][0],0,10))) {
$count++;
if($count % 2 == 0){
$in = '✔';
$out = '';
}
else{
$in = '';
$out = '✔';
}
}
else{
$count = 0;
$in = '✔';
$out = '';
}
}
However, I was instructed to add a break-in and break-out.
I'm using a hardware wherein it has a database on MS SQL. The types set are:
Type = 1 is In, Type = 2 is Out, Type = 4 is Break Out, and Type = 8 is Break In.
I'm trying to get it using this:
$query = 'SELECT ActualTime, Type FROM TA3.dbo.TimeLogs WHERE TimeLogs.EmployeeID='.$passed.' ORDER BY TimeLogs.ActualTime
$stmt = $conn->query($query);
$stmt->execute();
$arr = $stmt->fetchAll();
if (isset($_POST['Type'])
{
$Type = $_POST['Type'];
}
if($Type == 1)
{
$in = '✔';
$bin = '';
$bout = '';
$out = '';
}
if($Type == 2)
{
$in = '';
$bin = '';
$bout = '';
$out = '✔';
}
if($Type == 4)
{
$in = '';
$bin = '';
$bout = '✔';
$out = '';
}
if($Type == 8)
{
$in = '';
$bin = '✔';
$bout = '';
$out = '';
}
else
{
$in = '✔';
$bin = '';
$bout = '';
$out = '';
}
However, I'm not getting any checks on my table.