Question

SEE TABLE 表显示最终输出。所以我希望用户能够单击BlockName字段并单击BlockName上的数据应该发送到另一个页面。＆LT; /第＆GT; - ＆GT; 标题声母姓电话电子邮件办公室块名称部门照片 “;

                 $index = 0;

               $querynum ="SELECT csi315_dir_staff.title,csi315_dir_staff.initials,csi315_dir_staff.surname,csi315_dir_staff.phone,csi315_dir_staff.BlockName,csi315_dir_staff.email,csi315_dir_staff.office,csi315_dir_departments.department_name, csi315_dir_staff.photo FROM csi315_dir_staff, csi315_dir_departments WHERE csi315_dir_staff.department_id=csi315_dir_departments.department_id AND $search_query";
               $resultnum = mysqli_query($db_link,$querynum);

              $num_members = mysqli_num_rows($resultnum);
             // echo "search for  $search_name :" ;
              echo " $num_members staff  found. ";

                while ($row = mysqli_fetch_assoc($result)){
                $image="";
                if(file_exists("profile_pics/".$row['photo']))
                {
                    $image="profile_pics/".$row['photo'];
                }
                else{
                    $image="profile_pics/default.jpg";
                }   

                 echo "<tr  echo $index++ % 2 ? ' class='even'' : ''>",
                "<td>{$row['title']} </td>",
                "<td>{$row['initials']}</td>",
                "<td>{$row['surname']}</td>",
                "<td>{$row['phone']}</td>",
                "<td>{$row['email']}</td>",
                "<td>{$row['office']}</td>",
                "<td>{$row['department_name']}</td>",
                "<td><a href =\"/phonebook/withDropdownList_version2/home.html?id={$row['BlockName']}\"/><img src=\"images/build.png\" width=25px\" height=\"25px\"/></a>&nbsp;</td>",
                "</tr>";

            }
            echo "</tbody>
            </table>";

            // render the pagination links
            $pagination->render();`

Answer 1

您可以将数据作为查询参数发送。

如何在单击表行时将数据从PHP中的表行传递到另一个页面

1 个答案: