从填充的mysql下拉列表中插入选定的值

Question

我正在尝试从数据库中的引用表填充的下拉列表中插入所选值。 我遵循了有关动态下拉列表的教程，但现在我想取值并插入它。 问题是它一直在使用本教程使用的回显。 有什么方法可以使所选值成为新变量？ 当前它会插入“ <php echo $ team_name”

<div>
    <label>Home Team</label>
    <select name="home_team" style="width:125px;>
        <option value="">Select Team</option>
        <?php
            $query = "SELECT * FROM team";
            $results = mysqli_query($db, $query);
            mysqli_query($db, "SELECT * FROM team_name");
            // loop
            foreach ($results as $team_name) {
        ?>
        <option value="<php echo $team_name["cid"]; ?><?php echo $team_name["team_name"]; ?></option>
        <?php
        }
        ?>
   </select>

我如何尝试插入：

$db = mysqli_connect('localhost', 'root', 'root', 'register');
if(mysqli_connect_errno())
{
    echo "failed" . mysqli_connect_error();
}
//var_dump($_POST);
$home_team = mysqli_real_escape_string($db, $_POST['home_team']);
$home_team = $home_team;
$query = "INSERT INTO game_table (home_team)
          VALUES('$home_team')";
mysqli_query($db, $query);
//echo $query;
//echo $home_team;
//header('location: index.php');

Answer 1

请遵循此。

<select name="home_team" style="width:125px;>
    <option value="">Select Team</option>
    <?php
        $query = "SELECT * FROM team";
        $results = mysqli_query($db, $query);
        while($row = mysqli_fetch_assoc($results)) {
    ?>
    <option value="<php echo $row['cid']; ?>"><?php echo $row["team_name"]; ?></option>
    <?php } ?>
</select>

也许这应该工作

Answer 2

尝试这个。 您的代码中缺少两个"和“ ? " 。

<select name="home_team" style="width:125px;">
  <option value="">Select Team</option>
  <?php
    $query = "SELECT * FROM team";
    $results = mysqli_query($db, $query);
    foreach ($row = mysqli_fetch_assoc($results)) {
  ?>
      <option value="<?php echo $row["cid"]; ?>">
          <?php echo $row["team_name"]; ?>
      </option>
  <?php
    }
  ?>
</select>