PHP如何提交<select>從 mysql 獲取的選項值

Question

到目前為止，我正在獲取特定數據並將其顯示在選擇標記中作為我的代碼的選項：

   <?php
     $db = mysqli_connect('localhost', 'root', '', 'e-shop');
     $sql = 'SELECT * FROM categories';
     $result2 = mysqli_query($db,$sql);
   ?>
   <select name='category' >
        <?php while($row1 = mysqli_fetch_array($result2)):;?>
        <option value="<?php echo $row1[0];?>"><?php echo 
        $row1[1];?></option>
       <?php endwhile;?>
    </select>

我幾乎沒有上下文，我正在為我的書籍表單獲取類別，但我不知道如何將該值分配給類別

這是我提交的有效代碼：

<?php
  $db = mysqli_connect('localhost', 'root', '', 'e-shop');
  if (isset($_POST['add-product'])) {
    $title = $_POST['title'];
    $description = $_POST['description'];
    $price = $_POST['price'];
    $categoryy = $_POST['category'];
    $target_dir = "../img/";
    $fileName = basename($_FILES["image"]["name"]);
    $target_file = $target_dir . $fileName;
    move_uploaded_file($_FILES["image"]["tmp_name"], $target_file);
    $db_dir = "../img/";
    $db_file = $db_dir . $fileName;
    if($_FILES["image"]["name"]){
      $sql = sprintf("INSERT INTO products (title, category, picture, description, price) VALUES ('%s', '%s', '%s', '%s', '%s')", $title, $categoryy, $db_file, $description,  $price);
    } else {
      $sql = sprintf("INSERT INTO products (title, description, category, price) VALUES ('%s', '%s', '%s', '%s')", $title, $description,  $categoryy, $price);
    }
    mysqli_query($db, $sql);
    mysqli_close($db);
  }
?>

我想我需要一個名字來選擇標簽，我試過但沒有用，

Answer 1

顯然它確實有效，但我給了錯誤的行索引，該值是 id 而不是類別名稱！ 並命名標簽作品