第二个下拉php菜单不起作用php/mysqli/ajax

Question

更新：我得到的警告是''警告：mysqli_fetch_array（）期望参数1是mysqli_result，load_price.php中的布尔值15在第15行中给出''

我正在制作一个电子商务网站，我想下拉菜单，一个用于选择颜色（作品），一个用于价格（不起作用）

阅读了对价格不起作用的价格进行排序的相同步骤

菜单链接到外部PHP脚本文件，必须很简单外部文件：

<?php include "db/connect.php";?>
<?php
$output = '';
if(isset($_POST["price"]))
{
     if($_POST["price_id"] != '')
     {
          $sql = "SELECT * FROM 'clothing' ORDER BY 'price' '".$_POST["price_id"]."'";
     }
     else
     {
          $sql = "SELECT * FROM clothing";
     }
     $result = mysqli_query($conn, $sql);
     while($row = mysqli_fetch_array($result))
     {
          $output .= "
          <div class=items>
          <a href=viewitem.php?itemid=".$row['itemid'].">
          <div class='effect'>
          <img class='image' src=clothing/" . $row["firstimage"]. " class=img-responsive style=width:100% alt=Image>
          <img class='image hover' src=clothing/" . $row["secimage"]. " class=img-responsive style=width:100% alt=Image>
          </div>
          <div class=item-names>" . $row["productname"]. "</div>
          <div class=item-prices>$" . $row["price"]. "</div>
          </div>
          ";
     }
     echo $output;
}
?>
HTML file

<script>
  $(document).ready(function(){
       $('#price').change(function(){
            var price_id = $(this).val();
            $.ajax({
                 url:"load_price.php",
                 method:"POST",
                 data:{price_id},
                 success:function(data){
                      $('#show_product').html(data);
                 }
            });
       });
  });
  </script>
<?php
function fill_price($conn)
{
     $output = '';
     $sql = "SELECT * FROM price";
     $result = mysqli_query($conn, $sql);
     while($row = mysqli_fetch_array($result))
     {
          $output .= '<option value="'.$row["price_id"].'">'.$row["price"].'</option>';
     }
     return $output;
}
 //load_data_select.php
 function fill_color($conn)
 {
      $output = '';
      $sql = "SELECT * FROM color";
      $result = mysqli_query($conn, $sql);
      while($row = mysqli_fetch_array($result))
      {
           $output .= '<option value="'.$row["color_id"].'">'.$row["color"].'</option>';
      }
      return $output;
 }
 function fill_product($conn)
 {
      $output = '';
      $sql = "SELECT * FROM clothing";
      $result = mysqli_query($conn, $sql);
      while($row = mysqli_fetch_array($result))
      {
        $output .= "
        ";
   }
      return $output;
 }
 ?>
    
           <select name="price" id="price">
                <option value="">Show All</option>
                <?php echo fill_price($conn); ?>
           </select>
                     <select name="color" id="color">
                          <option value="">Show All Colours</option>
                          <?php echo fill_color($conn); ?>
                     </select>
                     <br /><br />
                     <div id="show_product">
                          <?php echo fill_product($conn);?>
                     </div>
</div>

Answer 1

语法错误。

因此，只需在单引号中添加classes。

<?php
$output = '';
while($row = mysqli_fetch_array($result))
  {
      $output .= "
      <div class='item-names'>" . $row["productname"]. "</div>
      <div class='item-prices'>$" . $row["price"]. "</div>
       ";
 }
 echo $output;
?>

图像标签（示例）

<td>
 <img class='product-image' src='http://localhost/product_loader/uploads/". $row["product_image_path"]. "'>
</td>

修复您的图像标签：

<img class='image img-responsive' src='clothing/". $row["firstimage"]. "'  style='width:100%;' alt='Image'>