Identifying time in & time out in a column - php

Question

I have a Library system for time in & out for the students. The students can go in when they want and go out when they want. This is how it works: There is a barcode on each student ID. They need to hover under the scanner every-time they pass the door, which automatically saves the Data.
Here is my code for saving.
<?php
$scan=$_GET["scan"];
include("connection/mysqlconnect.php");
$sql="SELECT * FROM student where Barcode like('$scan%')";
$result = $conn->query($sql);
$count=mysqli_num_rows($result);
if($count==1)
{
while($row = mysqli_fetch_array($result)) {
$id = $row['ID'];
$crs_id = $row['course_id'];
}
$my_date = date("Y-m-d h:i:s", strtotime("+7 hours"));
$sql="insert into dtr
(students_id,course_id,Date_Time)values('$id','$crs_id','$my_date')";
$result = $conn->query($sql) or die(mysqli_error());
mysqli_close($conn);
}
?>
The above code works perfectly.
But, I want to identify if it is in or out in a single column.
Below is my sample table. I want to add the In-Out Column.
Student_ID Course_ID Date_Time In-Out
11 4 2018-02-09 08:31:05 in
22 5 2018-02-09 09:35:09 in
22 5 2018-02-09 09:45:08 out
11 4 2018-02-09 10:01:05 out
22 5 2018-02-09 11:35:09 in
My problem is: I want to identify if the last hover is "in", in the column "In-Out" and the system will save the "out".

Answer 1

The simple way would be to select the records from the dtr table for that student/course combination and just fetch the last one...
$sql="SELECT `In-out` as lastAccess FROM `dtr` where Student_ID = $id
order by Date_Time desc limit 1";
$result2 = $conn->query($sql);
$count2=mysqli_num_rows($result2);
$lastAccess = 'out';
if( $count2 == 1 ) {
$row = mysqli_fetch_assoc($result2);
$lastAccess = $row['lastAccess'];
}
$lastAccess = ( $lastAccess == 'out' )? 'in':'out';
This will go after the first SQL to check the student and before the insert. The insert can then use $lastAccess.
Note that if a row in this table isn't found, the value of $lastAccess is 'out' (set prior to the test and therefore the bit which inverts this will set a new value of 'in' as it's the first time this student has used the system.

Answer 2

Here #Nigel
<?php
$scan=$_GET["scan"];
include("connection/mysqlconnect.php");
$sql="SELECT * FROM student where Barcode like('$scan%')";
$result = $conn->query($sql);
$count=mysqli_num_rows($result);
if($count==1)
{
while($row = mysqli_fetch_array($result)) {
$id = $row['ID'];
$crs_id = $row['course_id'];
}
$my_date = date("Y-m-d h:i:s", strtotime("+7 hours"));
$sql="SELECT `in-out` as lastAccess
FROM `dtr`
where student_id = $id
order by Date_Time desc
limit 1";
$result2 = $conn->query($sql);
$count2=mysqli_num_rows($result2);
$lastAccess = 'out';
if( $count2 == 1 ) {
$row = mysqli_fetch_assoc($result2);
$lastAccess = $row['lastAccess'];
// Check
echo "in-out=".$lastAccess.PHP_EOL;
}
$lastAccess = ( $lastAccess == 'out' )? 'in':'out';
// Check
echo "next in-out=".$lastAccess.PHP_EOL;
$sql="insert into dtr (students_id,course_id,Date_Time,in-out)values('$id','$crs_id','$my_date','$lastAccess')";
$result = $conn->query($sql) or die(mysqli_error($conn));
mysqli_close($conn);
}
?>