Question

目前，我从数据库中检索“01/05/2014”和“01/06/2014”等DATE记录，并显示为选择选项。

这是我的代码：

<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>

我获得的每年的产出将是2014年，2014年是重复的。如何才能在下拉列表中仅显示一个2014？

Answer 1

你可以像这样获得截然不同的年份，

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

（未经测试）

Answer 2

试试这个

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>';
 }

如何列出没有重复值的下拉选项

2 个答案: