php - 如何预选表单中的下拉选项和复选框选项(PHP/MySQL)？

Question

我正在创建一个更新表单，其中以前从 mysql 数据库提交的信息被抓取以填充当前表单。所以这就是我到目前为止所拥有的:

$select = mysql_query("SELECT * FROM some_table WHERE id = $id");
while ($return = mysql_fetch_assoc($select)) {
    $name = $return['name'];
    $bio = $return['bio'];
    $maritalStatus = $return['marital_status'];
    $favFood = $return['fav_food'];
}

<form action="page.php" method="post">
Name: <input type="text" name="name" value="<?php echo $name; ?>" /><br />

Bio: <textarea name="bio"><?php echo $bio; ?></textarea><br />

Marital Status
<select name="maritalStatus">
   <option>Select One</option>
   <option value="married">Married</option>
   <option value="single">Single</option>
   <option value="divorced">Divorced</option>
</select><br />

Favorite Food:
Cheeze: <input type="checkbox" name="favFood" value="cheeze" />
Cake: <input type="checkbox" name="favFood" value="cake" />
Oranges: <input type="checkbox" name="favFood" value="oranges" />
</form>

如您所见，我能够很好地显示通过输入文本框或文本区域输入的数据。但是，如果这两个选项是数据库中实际存在的选项，我该如何将“婚姻状况”下拉菜单预选为“离婚”选项并选中“最喜欢的食物”下的“橙子”复选框？

Answer 1

或者，为了避免向每个选项添加 if，只需执行以下操作:

echo '<select name="maritalStatus">';
$status = array("Select one", "Married", "Single", "Divorced");
foreach($status as $s)
      {
      $sel = ($maritalStatus == $s) ? 'selected = "selected"' : '';
      echo '<option value="'.$s.'" '.$sel.'>'.$s.'</option>';
      }
echo '</select>';

---

编辑:

要从数据库中动态填充选择，您可以:

echo '<select name="maritalStatus">';

$res = mysql_query("SELECT status FROM marital_status");
while ($row = mysql_fetch_array($res))
      {
      $s = $row['status']
      $sel = ($maritalStatus == $s) ? 'selected = "selected"' : '';
      echo '<option value="'.$s.'" '.$sel.'>'.$s.'</option>';
      }
echo '</select>';