PHP AJAX update MySql query onChange from下拉菜单（筛选）

Question

我想要的是：当用户从多个下拉菜单中选择不同的选项时，查询通过Ajax POST请求更新。因此，通过选择下拉列表来过滤结果。

HTML：

<select id="Gender">
    <option value="" disabled selected>--Geslacht--</option>
    <option value="1">Man</option>
    <option value="2">Vrouw</option>
    <option value="3">Beide</option>
</select>

选择数据的PHP MySQL查询：

$statements = '';
    if (isset($_POST['GenderID'])) {
         $gender_query = $_POST['GenderID'];
        $statements .= " AND `product`.`gender_id` = '$gender_query' "; //condition for each property
    }


    $query = "SELECT    `product`.`product_name` AS pname, 
                        `product`.`product_img` AS pimg,
                        `brand`.`brand_name` AS bname
            FROM
                        `product`                                           
            INNER JOIN
                        `brand`
            ON 
                        (`product`.`brand_id`=`brand`.`brand_id`)   
            WHERE 
                        `product`.`active` = '1' $statements";

  $result2 = $conn->query($query);  

                    if ($result2->num_rows > 0) {
                    // output data of each row
                       while($row = $result2->fetch_assoc()) 
                        {                       
                            echo'
                            <div class="col-lg-4">
                                <div class="card">

                                   <div class="view overlay">

                                    <a href="#">
                                        <img src="products/',$row["pimg"],'" class="img-fluid">
                                    </a>
                                        </div>

                                        <div class="card-body">
                                            <h4 class="card-title">',$row["bname"],'</h4>
                                            <p class="card-text">',$row["pname"],'</p>
                                            <a href="#" class="btn btn-outline-success my-2 my-sm-0">Button</a>
                                        </div>

                                    </div>
                                    </div>';
                                }
                            }

$(document).ready(function(){
    $("#Gender").change(function () {
        var gender = $("#Gender").val();
        jQuery.ajax({
            type: "POST",
            data:  {GenderID: gender},
            success: function(data){ 
                if(data.success == true){ 
                   alert('success'); 
                }
            }               
        });
    });
});

更新

我已经添加了执行Mysql查询的代码。

Answer 1

我想你在这里走错了。

genderid中'的一个小错误。但是您不会传递URL。

$(document).ready(function(){
    $("#Gender").change(function () {
        var gender = $("#Gender").val();
        jQuery.ajax({
            url: "www.some_url.com"    // Send the data with your url.
            type: "POST",
            data:  {'GenderID': gender},     // Here you have written as {GenderID: gender} , not {'GenderID': gender}
            success: function(data){ 
                if(data.success == true){ 
                   alert('success'); 
                }
            }               
        });
    });
});

您还更改了.而不是，如下所示。

echo' <div class="col-lg-4">
          <div class="card">
                 <div class="view overlay">
                    <a href="#">
                      <img src="products/'.$row["pimg"].'" class="img-fluid">
                    </a>
                 </div>
                 <div class="card-body">
                 <h4 class="card-title">'.$row["bname"].'</h4>
                 <p class="card-text">'.$row["pname"].'</p>
                 <a href="#" class="btn btn-outline-success my-2 my-sm-0">Button</a>
           </div>
       </div>
  </div>';