How to compare $typ = $POST[‘type’] with the leave type selected?

Leave Entry Form I am not able to find a way to compare the selected leave type $typ === 2 with the Sick leave id.

The $typ = $_POST[‘typ’] store the leave type obtained from a database such as Annual leave and Sick Leave. If I select the leave type from a form and it is a sick leave then I would just need to insert leave_date and resume_date in leave table otherwise if the selected leave type is annual then I need to insert the leave_date, resume_date in the leave table as well to update the leave_entitlement table just to reduce leave_entitlement balance

This is a excerpt from my form

 <form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>" >
 <div class="row form-group">     
  <div class="col-lg-12">
  <label>Leave Type</label>
  <select size="1" name="typ">
  <?php 

        $typs = mysqli_query($dbconn,"select * from type") or die('Query 
        failed: ' . mysqli_error());

        while ($line = mysqli_fetch_array($typs)) 
        { 
            echo "<option value='" . $line['typ_id'] . "'>"; 
            echo $line['typ_name'] . "</option>"; 
        } 
    ?> 
    </select>
  </div>
 </div>  
    </form>

This my submit

  if(isset($_POST['submit']))
 {
        $empnum = $_POST['empnum'];

    //     echo $empnum;
        $leave_date = $_POST['leave_date'];
        $resume_date = $_POST['resume_date'];

       $typ = $_POST['typ'];//leave type e.g Annual Leave, Sick Leave

    // this is a function to get the number of leave days from a form 
       Function GetNoOfDays($leave_date,$resume_date){
            $leave_dater = $leave_date;
            $resume_dater = $resume_date;
            $datediff = $resume_date - $leave_date;
            return round($datediff / (60 * 60 * 24));               
       }
          echo $leavetaken =  GetNoOfDays(strtotime($leave_date),strtotime($resume_date));

        //if a leave type is annual leave, $qry1 is excuted in the if body otherwise
        //if a leave type is sick leave $qry1 and qry2 in the else are excuted

        if($typ === 2) // 2 is a sick leave id if leave type is equal to 2 (sick leave) 
        {
            $qry1 = "INSERT INTO eleave VALUES(''," .
            "'$empnum','$leave_date','$resume_date'," .
            "'$typ')";

        }else{
            $qry1 = "INSERT INTO eleave VALUES(''," .
            "'$empnum','$leave_date','$resume_date'," .
            "'$typ')";
            $qry2 = "UPDATE eleave_entitlement SET annual_entitlement = annual_entitlement-$leavetaken WHERE emp_num = $empnum";

        }
            // execute query
            $added = mysqli_query($dbconn,$qry1).mysqli_query($dbconn,$qry2);

            // report results
            if(($added) != "")
            echo  "Record added successfully." . "<br>";
            else
            {
                echo "ERROR: Record could not be added<br>" . 
                    mysqli_error($dbconn);
            }
            // close connection
            mysqli_close($dbconn);
        }


 ?>

The problem is when I select Annual Leave or Sick Leave from the form my else part of my if statement always executed. Meaning that the leave_entitlement balance is always reduced even when I select Sick Leave

Answer

You use comparator === which will compare both value and data type. The data type of $typ is String so "2" === 2 will return false. If you want to use === you can cast $typ to integer by using (int) $typ or just change === to ==.

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Answer