Question

  1) when types selected from dropdown his category and subcategory should be shown 
   in dropdown,problem : when we select types in category it is showing all the category
   and all the subcategory of other types also
   please solve this problem as soon as possible.

   inventory_list.php
    <select name="typ_id" id="typ_id">
     <option>Select a Type</option>
      <?php
       $get_types = "select * from types";
      $run_types = mysql_query($get_types);
     while ($row_types=mysql_fetch_array($run_types)){
          $typ_id= $row_types['typ_id'];
         $typ_name = $row_types['typ_name'];
       echo "<option value='$typ_id'>$typ_name</option>";     
      }
    ?>

    <select name="cat_id" id="cat_id">
    <option>Select a Category</option>
   <?php
     $get_category = "SELECT * FROM category ";
     $run_category = mysql_query($get_category);
     while ($row_category=mysql_fetch_array($run_category)){
         $cat_id= $row_category['cat_id'];
         $cat_name = $row_category['cat_name'];
      echo "<option value='$cat_id'>$cat_name</option>";      
      }
    ?>
   </select>

Answer 1

尝试以下代码

<script>
 function get_category(type_id){
        $.ajax({
            type: "POST",
            url: "get_category.php",
            data: { type_id : type_id } 
        }).done(function(data){
            $("#category").html(data);
        });
  }
</script>

<select name="typ_id" id="typ_id" onchange="get_category(this)">
     <option>Select a Type</option>
      <?php
       $get_types = "select * from types";
      $run_types = mysql_query($get_types);
     while ($row_types=mysql_fetch_array($run_types)){
          $typ_id= $row_types['typ_id'];
         $typ_name = $row_types['typ_name'];
       echo "<option value='$typ_id'>$typ_name</option>";     
      }
    ?>

<div id='category'></div>

get_category.php

    <?php
         $get_category = "SELECT * FROM category where type_id = $_POST['type_id']";
         $run_category = mysql_query($get_category);
         echo '<select name="cat_id" id="cat_id">
        <option>Select a Category</option>';

         while ($row_category=mysql_fetch_array($run_category)){
             $cat_id= $row_category['cat_id'];
             $cat_name = $row_category['cat_name'];
          echo "<option value='$cat_id'>$cat_name</option>";      
          }
    echo '</select>';

?>

当从下拉列表中选择类型时，他的类别和子类别应显示在下拉列表中

1 个答案: