Question

我正在为我公司的销售构建基于php的Web应用程序。我正在创建一种功能，使销售人员能够将他的客户放入列表中以评估他的行为。所以我有一个客户的屏幕，每行我都有一个下拉菜单显示：

客户已属于的列表
将客户添加到其他现有列表的能力
将客户添加到新列表的能力

我的问题是选项c：我在下拉列表中创建了一个表单，以便为新列表名称提供输入。代码适用于除第一个以外的所有行！我的代码是：

 /// check if the customer belong to a list

    $query1 = "SELECT `cust_fav_type` FROM `customers_favorites` WHERE cust_fav_fs_id = '{$customer_fs_id}' AND cust_fav_user_id='{$_SESSION['user_id']}' ";
                $select_cb_lists = mysqli_query($connection,$query1);

                if(mysqli_num_rows($select_cb_lists)){

                    $count = mysqli_num_rows($select_cb_lists);

    /// if he exists I m creating a link with remove ability                
                    ?>
                    <td class="text-center">
                    <div class="dropdown">
                            <h4>
                                <button class="btn btn-info dropdown-toggle btn-xs" type="button" data-toggle="dropdown"> <b><?php echo $count;?></b>
                                <span class="caret"></span></button>
                                <ul class="dropdown-menu dropdown-menu-right">

                       <?php

                            while($row=mysqli_fetch_assoc($select_cb_lists)){

                            $c_list = $row['cust_fav_type'];



                            echo "<li><a href='posts.php?source=posts_list&cust_fs_id=$customer_fs_id&removefromlist=$c_list' onClick=\"javascript: return confirm('Σιγουρα θες να αφαιρέσεις τον $customer_name απο την λίστα $c_list?'); \"> <b>$c_list</b> &nbsp;<i class='text-danger'><span class='glyphicon glyphicon-remove'></span></i></a></li>";
                            }



/// ...and the ability to add to an existing list..

                                $query = "SELECT ul.cust_fav_type FROM users_lists ul LEFT JOIN (SELECT `cust_fav_type` FROM `customers_favorites` WHERE cust_fav_fs_id = '{$customer_fs_id}' AND cust_fav_user_id='{$_SESSION['user_id']}') as a ON ul.`cust_fav_type` =a.`cust_fav_type` WHERE ul.`cust_fav_user_id`='{$_SESSION['user_id']}' AND a.cust_fav_type IS NULL ";
                                $other_lists = mysqli_query($connection,$query);

                                if(mysqli_num_rows($other_lists)){

                                echo "<li class='divider'></li> ";

                                while($row=mysqli_fetch_assoc($other_lists)){

                                    $list_name = $row['cust_fav_type'];

                                    echo "<li><a href='posts.php?source=posts_list&cust_fs_id=$customer_fs_id&addto=$list_name'>Προσθηκη σε <b>$list_name</b></a> </li> ";


                                }
                                }

                            ?>        
                   <br>

    /// THE Problem!! a form for inserting customer to a new list


                     <li>
                            <form action="posts.php" method="get">
                                <div class="input-group">
                                    <input class="form-control" type="hidden" name="source" value="posts_list">
                                    <input type="hidden" name="cust_fs_id" value="<?php echo $customer_fs_id;?>">                           
                                    <input class="form-control" type="text" name="addto" placeholder="Προσθηκη σε νεα λίστα">
                                    <span class='input-group-btn'> 
                                    <button class='btn btn-primary' type="submit"><span class='glyphicon glyphicon-plus'></span></button></span>
                                </div>
                            </form>
                        </li>    

                    </ul>



                    </h4>
                    </div>

                </td>



                    <?php


                } else {

                    $count = "-";


                    ?>

                    <td class="text-center">
                        <div class="dropdown"><h4>
                                <button class="btn btn-primary dropdown-toggle btn-xs" type="button" data-toggle="dropdown"><span class='glyphicon glyphicon-plus'></span><span class="caret"></span></button>
                                <ul class="dropdown-menu dropdown-menu-right">

                           <?php

                                $query = "SELECT cust_fav_type FROM customers_favorites ";
                                $query .= " WHERE cust_fav_user_id={$_SESSION['user_id']} ";
                                $query .= "GROUP BY cust_fav_type ORDER BY cust_fav_type ";
                                $select_customers_cities = mysqli_query($connection,$query);

                                while($row = mysqli_fetch_assoc($select_customers_cities)) {
                                $customers_filters = $row['cust_fav_type'];


                                echo "<li><a href='posts.php?source=posts_list&cust_fs_id=$customer_fs_id&addto=$customers_filters'>Προσθήκη σε <b>$customers_filters</b></a></li>";
                                }

                                ?>
                     </ul>
                                </h4>
                                </div>     

                      </td> 


                    <?php  } ?>

正如我写的那样，除了第一个以外，所有客户都能正常工作！任何帮助都会有价值！

感谢名单！

表单不适用于第一条记录

0 个答案: