Question

  <?php
      $sql = "SELECT id, name FROM cities";
      $stmt = $DBcon->prepare($sql);
      $stmt->execute();
      $cities = $stmt->fetchAll();

      $current_city_id = '';

      $current_city_id=$cities[0]['id'];
      $dropdown = '<select name="city">';
      foreach ($cities as $city) {
        $selected='';
        if ($current_city_id==$city['id']) $selected=" selected";

        $dropdown .= '<option value="'.$city['id'].'">' . $city['name'] .'</option>';

      }
      $dropdown .='</select>';
      echo $dropdown;
      ?>

选择的值未显示在下拉菜单中，然后用户按下更新按钮。无法找到问题所在。

Answer 1

您没有在选项...

中添加 $ selected 值

<?php
    $sql = "SELECT id, name FROM cities";
    $stmt = $DBcon->prepare($sql);
    $stmt->execute();
    $cities = $stmt->fetchAll();
    $current_city_id = '';
    $current_city_id=$cities[0]['id'];
    $dropdown = '<select name="city">';
    foreach ($cities as $city) {
        $selected='';
        if ($current_city_id==$city['id']) $selected=" selected";
        $dropdown .= '<option value="'.$city['id'].'" '.$selected.'>' . $city['name'] .'</option>'; //Change Here  
    }
    $dropdown .='</select>';
    echo $dropdown;
?>

Answer 2

您不打印$ selected变量。 $ selected变量应打印在选项

中

 <?php
          $sql = "SELECT id, name FROM cities";
          $stmt = $DBcon->prepare($sql);
          $stmt->execute();
          $cities = $stmt->fetchAll();

            $current_city_id = '';

          $current_city_id=$cities[0]['id'];
          $dropdown = '<select name="city">';
          foreach ($cities as $city) {
            $selected='';
          if ($current_city_id==$city['id']) $selected=" selected";
    //print out selected variable
          $dropdown .= '<option '.$selected.' value="'.$city['id'].'">' . $city['name'] .'</option>';

        }
          $dropdown .='</select>';
          echo $dropdown;
          ?>

Php select =选择不起作用

2 个答案: