使用完整路径 (URLS) 和数据库中的值生成嵌套树
我从昨天开始就在研究这个问题。更具体地说 - 我在数据库中有一些值,如下所示:
____________________________________ | id | parentId | name | url | ------------------------------------ | 1 | 0 | Nieuws | url_1 | ------------------------------------ | 2 | 0 | Reviews | url_2 | ------------------------------------ | 3 | 0 | Meuk | url_3 | ------------------------------------ | 4 | 1 | Games | url_4 | ------------------------------------ | 5 | 1 | Internet | url_5 | ------------------------------------ | 6 | 5 | Browsers | url_6 | ------------------------------------
我希望看到基于这些值生成的树。目标形状应如下所示:
<ul>
<li><a href="url_1">Nieuws</a>
<ul>
<li><a href="url_1/url_4">Games</a></li>
<li><a href="url_1/url_5">Internet</a>
<ul>
<li><a href="url_1/url_5/url_6">Browsers</a></li>
</ul>
</li>
</ul>
</li>
<li><a href="url_2">Reviews</a></li>
<li><a href="url_3">Meuk</a></li>
</ul>
对我来说,将所有斜线包含在树内(包含所有父级和子级的完整路径)非常重要。
我想补充一点,我在此页面上找到了代码: http://crisp.tweakblogs.net/blog/317/formatting-a-multi-level-menu-using-only-one-query.html 但我可以不是按照上述方式重做。我将非常感谢任何帮助,因为申请截止日期即将到来:(
I work on this issue since yesterday. More specifically - I have some values in the database, which look as follows:
____________________________________ | id | parentId | name | url | ------------------------------------ | 1 | 0 | Nieuws | url_1 | ------------------------------------ | 2 | 0 | Reviews | url_2 | ------------------------------------ | 3 | 0 | Meuk | url_3 | ------------------------------------ | 4 | 1 | Games | url_4 | ------------------------------------ | 5 | 1 | Internet | url_5 | ------------------------------------ | 6 | 5 | Browsers | url_6 | ------------------------------------
I would like to see generated tree on the basis of these values. Target shape should look like this:
<ul>
<li><a href="url_1">Nieuws</a>
<ul>
<li><a href="url_1/url_4">Games</a></li>
<li><a href="url_1/url_5">Internet</a>
<ul>
<li><a href="url_1/url_5/url_6">Browsers</a></li>
</ul>
</li>
</ul>
</li>
<li><a href="url_2">Reviews</a></li>
<li><a href="url_3">Meuk</a></li>
</ul>
It's important for me to have all the slashes inside the tree (full path with all the parents & children).
I would like to add, that I found the code on this page: http://crisp.tweakblogs.net/blog/317/formatting-a-multi-level-menu-using-only-one-query.html but I can not redo it in such a manner as described above. I'll be very grateful for any help, because the application deadline is approaching very fast :(
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仅一处修改。我觉得这个方法简单多了。我没有使用数组来存储 URL,而是传递一个字符串作为参数(包含基本 url)。
Only one modification. I think this way is much simpler. Instead of using an array to store the URL, I pass a string as an argument (containing the base url).
根据链接中的函数,如果将 url 添加到查询中,类似这样的操作应该有效:
based on the function in your link, if you add url to the query, something like this should work: