Question

我正在尝试创建一个动态下拉列表，根据下拉列值显示输出。我一直在尝试运行此代码，但我收到此警告。我不知道为什么。 警告：mysqli_fetch_assoc（）要求参数1为mysqli_result，第18行的C：\ wamp64 \ www \ testing \ load_data.php中给出布尔值

      <?php
    define("DB_HOST", "localhost");
    define("DB_USER", "root");
    define("DB_PASS", "");
    define("DB_NAME", "dd_vals");

    $connection = mysqli_connect(DB_HOST, DB_USER, DB_PASS, DB_NAME);

    if(!$connection) {
        echo "connection failed";
    }
?>

<!DOCTYPE html>
<html>
<head>
    <script src="http://ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"></script>
</head>
<body>

    <select id="category">

        <?php

            $sql = "select * from dd_vals";
            $result = mysqli_query($connection,$sql);
            while($row = mysqli_fetch_assoc($result)){
                $index = $row['index'];
                $category = $row['category'];
        ?>
                <option value="<?php echo $index; ?>"><?php echo $category; ?> 
        </option>
        <?php }  ?>


       </select>


       <div id="stuff"></div>



      <script>  
      $(document).ready(function(){
        $('#category').change(function(){
                var index = $(this).val();
            $('#stuff').load("load_data.php", {
                index : index
            });
        });

      });
      </script>   

      </body>
      </html>

这是加载数据文件

     <?php
  define("DB_HOST", "localhost");
  define("DB_USER", "root");
  define("DB_PASS", "");
  define("DB_NAME", "dd_vals");

  $connection = mysqli_connect(DB_HOST, DB_USER, DB_PASS, DB_NAME);

  if(!$connection) {
    echo "connection failed";
  }

$index = $_POST['index'];

$sql = "select * from dd_vals where index = $index LIMIT 1";
$result = mysqli_query($connection, $sql);

while($row = mysqli_fetch_assoc($result)){
    echo $row['dd_val']; 
}

?>

我正在尝试创建动态下拉列表

0 个答案: