困惑使Sitemap使用PHP编码 [英] confused to make sitemap using php coding

问题描述

我已经创建了一个CMS网站，现在我试图使php编码的网站地图，但我所做的代码是不完整的。即它不显示所有的子菜单数据。

请观看此图片。根据这个图像，我有很多其他子菜单下重量训练（子菜单的健身运动），但他们是不可见的。

请指导/帮助我解决此问题

我的php代码

p>

  foreach（$ query_sitemap as $ artrow）{
 $ datefromsql = $ artrow-> created_date; 
 $ time = strtotime（$ datefromsql）; > 
 
< p> 
< ul> 
<？php echo'< b>'。$ artrow-> menu_name。< / b>'; > 
 
 
<？php $ submenucon = $ this-> menumodel-> fetch_menu_byPid（$ artrow-> id）; // ie select * from menu where parent_id = $ mid 
 if（empty（$ submenucon））{？> 
 --- 
<？php 
} else {
 foreach（$ submenucon as $ subtrow）{
？ 
<？php echo'< li style =color：grey; margin：>'。$ subtrow-> menu_name。< / li>< br / > 
<？php 
} 
} 
？> 
< / ul> 
< / p> 
<？php 
 $ i ++; 
} 
}？> 
  

解决方案

在最普通/朴素的方式）在数据库。因此，你应该做的第一件事是建立一个基于返回的行的树状结构，然后再遍历它为了显示。

你可以阅读更多在这里解析树状表格：

p>

实施分层数据库中的数据结构

将平面表解析成树的最有效/优雅的方法是什么？

I have created a CMS website, now I am trying to make site map with php coding, but the code which i have made is not complete. i.e It is not showing all the sub menu data.

Please look at this image . According to this image, I am having many other sub menu under Weight training(sub menu of Fitness Exercises) ,but they are not visible. Why they are not visibile?

Please guide/help me to solve this issue

My php code

   foreach ($query_sitemap as $artrow) {
        $datefromsql = $artrow->created_date;
          $time = strtotime($datefromsql);  ?> 

                <p>    
                        <ul>
                            <?php echo '<b>'.$artrow->menu_name.'</b>'; ?>


                                <?php $submenucon=$this->menumodel->fetch_menu_byPid($artrow->id);//i.e select*from menu where parent_id=$mid
                                 if (empty($submenucon)) { ?>
                                  ---
                                 <?php
                                    } else {
                                         foreach ($submenucon as $subtrow) {
                                             ?>
                                              <?php echo '<li style="color:gray;margin:">'.$subtrow->menu_name.'</li><br/>'; ?>
                                      <?php
                                         }
                                    }
                                 ?>    
                        </ul> 
                        </p> 
                        <?php
                        $i++;
                    }
                }    ?> 

解决方案

You're dealing with a tree structure here represented (in a most siplistic/naive way) in a database. So the first thing you should do is to build a tree-like structure based on returned rows and only then traverse it recrusively in order to display.

You can read more on parsing tree-like tables here:

Is it possible to query a tree structure table in MySQL in a single query, to any depth?

Implementing a hierarchical data structure in a database

What is the most efficient/elegant way to parse a flat table into a tree?

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