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

Question

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;
}
}

Answer 1

Assuming your date of birth field is a MySQL DATE field, you need to make sure that you are creating a date in the correct format i.e. YYYY-MM-DD. You can do that by using strtotime to convert your input data into a timestamp, and then date to convert that into the appropriate format:
$DateOfBirth = date('Y-m-d', strtotime("$date $month $year"));
For example:
$year = 2015;
$month = 'August';
$date = 5;
$DateOfBirth = date('Y-m-d', strtotime("$date $month $year"));
echo $DateOfBirth;
Output:
2015-08-05
Demo on 3v4l.org
Note
If you really want the result to be in dd/mm/yyyy format, just change the first input to date to 'd/m/Y'.

Answer 2

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