CODEWITHSUNDEEP
php
user3663143
user3663143

Reputation: 63

How to list out dropdown option without duplicate value

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?

Upvotes: 0

Views: 830

Answers (2)

krishna
krishna

Reputation: 4089

try this

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;
     // save year into an array 
     $yr[] = $year;
     // echo only if year is not in array
     if(!in_array($year,$yr))
     echo '<option>'.$year.'</option>';
 }

Upvotes: 0

Subir Kumar Sao
Subir Kumar Sao

Reputation: 8411

You can get distinct year like this,

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

(Not tested)

Upvotes: 2

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