尝试将Selected Attribute添加到while语句填充的下拉列表中，该语句使用mysql中的数组 - Trying to add Selected Attribute to dropdown populated by while statement with array from mysql

Trying to add Selected Attribute to dropdown populated by while statement with array from mysql

我有一个带有下拉菜单的表单，它由while语句填充，从mysql中的数据库中提取一个数组。我正在尝试为其中一个选项添加一个选定的属性，该属性已存储在页面的其他地方的变量$inv_item["type"]中

function find_all_types() {
    global $connection;
    $query  = "SELECT type ";
    $query .= "FROM types ";
    $query .= "ORDER BY id ASC";
    $type_set = mysqli_query($connection, $query);
    confirm_query($type_set);
    return $type_set;
}
       <select name="type">
           <?php
               $type_set = find_all_types();
               while ($type = mysqli_fetch_assoc($type_set)){ 
           ?>
           <option value= "<?php echo $type['type'] ?>"><?php echo $type['type'] ?></option>
           <?php } ?>
       </select>

我不确定是否有更好的方法来实现这一点，或者我是否可以使用while语句来实现它。这很难研究，我找不到同样的情况，大多数搜索结果的人都在用单独编写的选项填充下拉菜单。

我知道mysql命令是不推荐使用的，正在处理这个问题。

you could do it as a comparison in the option:
<option value= "<?php echo $type['type'] ?>" <?php if($type['type'] == $inv_item["type"]){echo" selected";}?>><?php echo $type['type'] ?></option>

但它看起来很乱——进出php——我会留在里面：

        <?php
               $type_set = find_all_types();
               while ($type = mysqli_fetch_assoc($type_set)){ 
          $type_item = $type['type'];
          echo"<option value= '".$type_item."'";
           if($type_item == $inv_item["type"]){echo" selected";}
         echo" >".$type_item."</option>";
           } 
?>