Jquery on change function pass value using ajax to database file.但只第一次工作答案

【问题标题】：Jquery on change function pass value using ajax to database file .but working only first timeJquery on change function pass value using ajax to database file.但只第一次工作
【发布时间】：2018-01-04 07:48:57
【问题描述】：

Jquery On Change 函数仅在我使用 ajax jquery 函数将值传递给使用选择框的数据库文件时才首先工作。但是当我更改所选框的值时，它仅在第一次第二次工作时才工作。它不起作用为什么请帮助代码是下面给出。

if(isset($package_sort ) && !empty($package_sort ))
 {




switch ($package_sort) {

    case "1000":
       $sql = "SELECT * FROM package WHERE package_city='goa' AND package_price ORDER BY package_price ASC";   

        break;
    case "160000":
        $sql = "SELECT * FROM package WHERE package_city='goa' AND package_price <=".$_GET['package_sort']." ORDER BY package_price DESC";   

        break;
    default:
        echo "Your favorite color is neither red, blue, nor green!";
}

<script>  

    $(function(){  
   $(document).on('change', '#package_sort', function(){  
           var package_sort = $(this).val();


           $.ajax({  
                url:"package-search.php",  
                method:"GET",  
                data:{package_sort:package_sort},  
                success:function(data){  
                     $("#package_loading").fadeIn(500).html(data);  
                }  

           });  
      });  
    });
    </script>  
<div class="select-wrapper"><select name="package_sort1" id="package_sort" class="custom-select selectbox">
                                                        <option value="" disabled="disabled" selected="selected" hidden="hidden">Sort By</option>
                                                        <option value="1000" name="package_sort">Low to Height</option>
                                                        <option value="160000" name="package_sort">Height to Low</option>

                                                    </select></div>

【问题讨论】：

试试这个$(document).ready(function(){ $("#package_sort").change(function(){ alert("change"); }); });检查警报。

标签： php jquery mysql ajax filtering

【解决方案1】：

您尚未定义变量$package_sort。您还需要检查$_GET 是否设置

if(isset($_GET['package_sort']) && !empty($_GET['package_sort'] )){
    $package_sort = $_GET['package_sort'];
    switch ($package_sort) {

        case "1000":
           $sql = "SELECT * FROM package WHERE package_city='goa' AND package_price ORDER BY package_price ASC";   

            break;
        case "160000":
            $sql = "SELECT * FROM package WHERE package_city='goa' AND package_price <=".$_GET['package_sort']." ORDER BY package_price DESC";   

            break;
        default:
            echo "Your favorite color is neither red, blue, nor green!";
    }
}

【讨论】：

我还将值 $_GET['package_sort'] 存储在 $package_sort 中
jquery 代码 -