如何列出没有重复值的下拉选项

Question

Currently I retrieving DATE records like "01/05/2014" and "01/06/2014" from database and display into select option.

Here my code:

<label>Year</label>
   <select style="margin-right:20px;">
       <option value="--" selected="selected">--</option>
       <?php
         require('Retail/database_connection.php');
         if (@mysqli_connect_errno())
         {
            echo "Failed to connect to MySQL: " . mysqli_connect_error();
         }

        $select_album_date= "SELECT DISTINCT photo_album_date FROM tblPhotoAlbum";
        $result = mysqli_query ($mydatabase, $select_album_date);
        if($result)
        {
           while($row = mysqli_fetch_array($result))
           {
            $date = $row['photo_album_date'];

            $arr = explode("/", $date);
            //Split date result into $day, $month, and $year
            list($month, $day, $year) = $arr;


            echo '<option>'.$year.'</option>';



           }
        }mysqli_close($mydatabase);
      ?>
  </select>

The output of $year I get will be 2014, 2014 which is duplicated. How can I only display only one 2014 in my dropdown list?

Answer 1

You can get distinct year like this,

SELECT DISTINCT YEAR(STR_TO_DATE(photo_album_date,'%d/%m/%Y')) FROM tblPhotoAlbum

(Not tested)