选中状态mysql数组

Question

我正在尝试将表选择输出与我发现的一段代码结合起来以保持所选状态：

<select>
<?php
$desired_option = 'arsenal';
$arr = array('arsenal', 'aston villa', 'birmingham', 'blackpool', 'bolton');
for($i = 0; $i < count($arr); $i++) {
$selected = ($arr[$i] == $desired_option) ? 'selected="selected"' : '';
echo "<option value=\"{$arr[$i]}\" {$selected}>{$arr[$i]}</option>";
}
?>
</select>

<select id="teams" onchange="this.form.submit();" name="teamid">
<? 
include('db.php');
$getTeams = mysql_query("SELECT name, id FROM team") or die(mysql_error());
while ($teamsData = mysql_fetch_array($getTeams))
{
?>    
<option value="<? echo $teamsData['id']; ?>" ><? echo $teamsData['name']; ?></option>
<? 
}
?>
</select>

我已经尝试了一切。有什么想法吗？

谢谢：）

Answer 1

<select id="teams" onchange="this.form.submit();" name="teamid">
<? 
include('db.php');

$selected = 'team_to_be_selected';

$getTeams = mysql_query("SELECT name, id FROM team") or die(mysql_error());
while ($teamsData = mysql_fetch_array($getTeams))
{
?>    
<option value="<?php echo $teamsData['id']; ?>" <?php echo ($teamsData['name'] == $selected) ? 'selected="selected"' : ''; ?>><?php echo $teamData['name'];?></option>
<? 
}
?>
</select>