如何使用php,mysql将单击列表项放入导航栏?

时间:2018-10-20 09:06:07

标签: javascript php html mysql

我想在导航栏中显示所选或单击的产品类型的品牌名称。 SQL代码可以,因为我已经测试过了。我不知道php有什么问题。我试图知道用Javascript选择了哪个项目。

<ul class="nav navbar-nav ">
   <li class="active">
      <a href="index.php">Home</a>
        </li>
   <li>
      <a href="#" class="dropdown-toggle" data-toggle="dropdown">Products <span class="caret"></span></a>
            <ul class="dropdown-menu multi-level" id="productNav">
                                                <?php
            $get_p_cats = "select * from category";
            $run_p_cats = mysqli_query($con,$get_p_cats);
            $selectedProductID="<script type='text/javascript'>
                                                var ul = document.getElementById('productNav');
                                                ul.onclick = function(event) {
                                                    var target = event.target;
                                                    alert(event.target.innerHTML);
                                                };  </script>";

                                                echo "<script type='text/javascript'>
                                                var ul = document.getElementById('productNav');
                                                ul.onclick = function(event) {
                                                    var target = event.target;
                                                    alert(event.target.innerHTML);
                                                };  </script>";

                                                $get_specificBrand="select distinct i.brandId,b.brandName 
                                                from brand b, item i 
                                                where b.brandId=i.brandId and i.categoryId=$selectedProductID";
            $run_specificBrand = mysqli_query($con,$get_specificBrand);
            while($row_p_cats = mysqli_fetch_array($run_p_cats) ){
            $p_cat_id = $row_p_cats['catId'];
            $p_cat_title = $row_p_cats['catName'];
            echo "<li class='dropdown-submenu'><a  class='dropdown-toggle' data-toggle='dropdown' href='shop.php?p_cat=$p_cat_id'> $p_cat_title </a>
                                                    <ul class='dropdown-menu'>
                                                    <li><a href='#'>p_specificBrandtitle</a></li></ul></li>";
            }
            ?>
            </ul>
                                        </li></ul>

0 个答案:

没有答案