[ACCEPTED]-Populate select drop down from a database table-mysql

Accepted answer

Score: 18

$query = "SELECT volunteers_2009.id, volunteers_2009.comments, volunteers_2009.choice1, volunteers_2009.choice2, volunteers_2009.choice3, volunteers_2009.lname, volunteers_2009.fname, volunteers_2009.venue_id, venues.venue_name FROM volunteers_2009 AS volunteers_2009 LEFT OUTER JOIN venues ON (volunteers_2009.venue_id = venues.id) ORDER by $order $sort";

$res = mysql_query($query);
echo "<select name = 'venue'>";
while (($row = mysql_fetch_row($res)) != null)
{
    echo "<option value = '{$row['venue_id']}'";
    if ($selected_venue_id == $row['venue_id'])
        echo "selected = 'selected'";
    echo ">{$row['venue_name']}</option>";
}
echo "</select>";

Score: 4

assuming you have an array of venues...personally 1 i don't like to mix the sql with other wizardry.

function displayDropDown($items, $name, $label, $default='') {
  if (count($items)) {
    echo '<select name="' . $name . '">';
    echo '<option value="">' . $label . '</option>';
    echo '<option value="">----------</option>';
    foreach($items as $item) {
      $selected = ($item['id'] == $default) ? ' selected="selected" : '';
      echo <option value="' . $item['id'] . '"' . $selected . '>' . $item['name'] . '</option>';
    }
    echo '</select>';
  } else {
    echo 'There are no venues';
  }
}

Score: 2

        <?php 
        $query = "SELECT * from blogcategory";
        //$res = mysql_query($query);
        $rows = $db->query($query);
        echo "<select name = 'venue'>";
        // while (($row = mysql_fetch_row($res)) != null)
        while ($record = $db->fetch_array($rows)) 
        {
            echo "<option value = '{$record['CategoryId']}'";
            if ($CategoryId == $record['CategoryId'])
                echo "selected = 'selected'";
            echo ">{$record['CategoryName']}</option>";
        }
        echo "</select>";
        ?>

Source: stackoverflow.com

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