选中状态mysql数组
我正在尝试将表选择输出与我发现的一段代码组合以保持所选状态：
<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>
没有找到相关结果
已邀请:
1 个回复
谦响局豢报
<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>