Question

您好我正在使用ajax来调用on-change下拉列表并从数据库中选择所有值到表行格式，但它只显示从数据库到表行的最后一条记录。我在数据库中有三行与下拉列表中的一个选择相关，我想将所有三行显示为表行格式。我的代码是

 $(document).ready(function(){
        $('#location').change(function(){

            $.ajax({
      url:'location.php?location=' + $(this).val(),
      type: "get",
      timeout:10000,
     dataType: 'json',
      success: function(data){

   document.getElementById("material_name").value=data.material_name;

   document.getElementById("material_quantity").value=data.product_quantity;

      },
      error:function(){

      }   
    });             
        });
and location.php page is as:-
    <?php
require_once("shine_class.php");
                $s = new shine;
                $s->connection();
$location = $_REQUEST['location'];
$query ="SELECT * from material_used where product_name = '$location'  ";
$query1=mysql_query($query);
while($row = mysql_fetch_array($query1))
{
 $result1=$row['material_name'];
 $result2=$row['product_quantity'];
}
 echo json_encode(array("material_name"=>$result1,"product_quantity"=>$result2));
?>

Ajax下拉调用函数无法显示所有值

0 个答案: