从数据库表中填充选择下拉列表

Question

我有一张桌子("场地"),存放志愿者可以工作的所有场地,每个志愿者被分配到一个场地工作.

我想从场地表创建一个选择下拉列表.

现在我可以显示每个志愿者分配的场地,但是我希望它显示下拉框,并且已经在列表中选择了场地.

<form action="upd.php?id=7">
<select name="venue_id">
<?php //some sort of loop goes here
print '<option value="'.$row['venue_id'].'">'.$row['venue_name'].'</option>';
//end loop here ?>
</select>
<input type="submit" value="submit" name="submit">
</form>

例如,id为7的志愿者被分配到venue_id 4

<form action="upd.php?id=7">
<select name="venue_id">
    <option value="1">Bagpipe Competition</option>
    <option value="2">Band Assistance</option>
    <option value="3">Beer/Wine Pouring</option>
    <option value="4" selected>Brochure Distribution</option>
    <option value="5">Childrens Area</option>
    <option value="6">Cleanup</option>
    <option value="7">Cultural Center Display</option>
    <option value="8">Festival Merch</option>
</select>
<input type="submit" value="submit" name="submit">
</form>

Brochure Distribution option will already be selected when it displays the drop down list, because in the volunteers_2009 table, column venue_id is 4.

我知道它将采用for或while循环的形式从场地表中拉出场地列表

我的查询是:

$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";

如何从场地桌面填充场地(volunteers_2009.venue_id,venues.id)的选择下拉框并让它预先选择列表中的场地？

Answer 1

$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>";

:)