How to add 3 drop down in single variable in php?

Jack Tech :

Two days ago I had an interview as a php developer and they gave me a task to perform. I completed 90% of the task, but I failed while trying to add date, month and year together.

I have 3 dropdowns for date, month and year:

Date Of Birth : <br />
Date : <select name="date">
            <?php $i=1; for($i=1; $i<=31; $i++){ ?> <option> <?php echo $i?> </option> <?php } ?>    
       </select>  
Month :  <select name="month">
            <option value="January">January</option>
            <option>Febuary</option>
            <option>March</option>
            <option>April</option>
            <option>May</option>
            <option>June</option>
            <option>July</option>
            <option>August</option>
            <option>September</option>
            <option>October</option>
            <option>November</option>
            <option>December</option>

        </select>
Year :    
<?php
    $currently_selected = date('Y'); 
    $earliest_year = 1950; 
    $latest_year = date('Y'); 

    print '<select name="year">';
    foreach ( range( $latest_year, $earliest_year ) as $i ) {
    print '<option value="'.$i.'"'.($i === $currently_selected ? ' selected="selected"' : '').'>'.$i.'</option>';
    }
    print '</select>';
?>

I have only one column in the database for the date of birth. I want to add the date, month and year together, which should result in the following: date/month/year. How could I accomplish this?

if(isset($_POST['submit'])){
    $name = trim($_POST['username']);
    $date = $_POST['date'];
    $month = $_POST['month'];
    $year = $_POST['year'];

    if($name == ''){
        $error = "Add Name";
    }    

    $DateOfBirth = $date.'/'$month;

    if(!$error){
        echo $DateOfBirth;
    }
}

Nipun Tharuksha :

The reason is your are not concatenating the variables in php. Check below code. You are almost there but with small coding mistake.

What you are trying to do is concatenate a variable with a string. So you have to concatenate variable with a '.' and then add your string. In here / works as the string. When adding a string it should be with in quotations '/' . And the final output will be something like this. $variable . '/'. Now concatenate another variable. Then it would be like this. $variable01 . '/' . $variable02.

 $DateOfBirth = $date.'/'.$month.'/'.$date ;

Output

12/02/2020

Check the added example with respect to your requirement

example