如何用更多字段填充数据库中的选择选项菜单 - How can I fill up select option menu from database with more fields?

How can I fill up select option menu from database with more fields?

我有一个选择选项菜单。我从数据库中填写：

$value = mysql_query("SELECT title FROM movies ORDER BY id DESC");
$i = 0;
while ($row = mysql_fetch_object($value))
{
    $release[$i] = $row->title;
    $i++;
}

我把它们写出来：

<select name="release"><?php 
foreach($release as $i){
    echo"<option value = '$i' > $i </option>";
}
?></select>

它运行良好，但我想在菜单中的标题旁边写下发布年份。上映年份在同一个电影表中，在上映字段中。如何追加这两个字段的值？

谢谢！

从数据库中选择两列，并将整行对象放入数组中：

$value = mysql_query("SELECT title, release_year FROM movies ORDER BY id DESC");
while ($row = mysql_fetch_object($value))
{
    $release[] = $row;
}

然后把它们都写出来：

<select name="release"><?php 
foreach($release as $i => $r){
    echo"<option value = '$r->title' > $r->title ($r->release_year) </option>";
}
?></select>

将代码更改为此

$value = mysql_query("SELECT title, release FROM movies ORDER BY id DESC");
$i = 0;
while ($row = mysql_fetch_object($value))
{
    $release[$i] = $row->title . ' (' . $row->release . ')';
    $i++;
}

试试这个：TESTED

    <select name="release">
        <?php
        $value = "SELECT title, release_year FROM movies ORDER BY id DESC";
        $select=mysql_query($value);
while($row=mysql_fetch_array($select))
    {
        $title=$row['title'];
      $release_year=$row['release_year'];
     echo $title_rel_yr=$title.'('.$release_year.')';
            echo"<option value = '$title' >$title_rel_yr</option>";
                }       
        ?>
      </select>