Question

我有一个Json菜单，我喜欢它： -

[{"slug":"index.php","name":"Home"},{"slug":"aboutus","name":"About us","children":[{"slug":"eims","name":"Eims"},{"slug":"vision","name":"Vision"}]},{"slug":"trash","name":"Trash","children":[{"slug":"ways","name":"Ways"},{"slug":"requ","name":"Requ"}]}]

现在我需要将这个json转换为这样的菜单： -

<li><a href="./"  >Home</a></li>
<li><a href="container.php?fun=aboutus" >About us</a>       
            <ul>
                <li><a href="container.php?fun=eims">Eims</a></li>
                <li><a href="container.php?fun=vision">Vision</a></li>
            </ul>
        </li>
<li><a href="container.php?fun=trash">Trash</a>
<ul>
                <li><a href="container.php?fun=ways">Ways</a></li>
                <li><a href="container.php?fun=requ">Requ</a></li>
            </ul>
        </li>
</li>

我该怎么做？

Answer 1

检查一下。

<?php
    function parseNodes($nodes) {
            $ul = "<ul>\n";
            foreach ($nodes as $node) {
                    $ul .= parseNode($node);
            }
            $ul .= "</ul>\n";
            return $ul;
    }

    function parseNode($node) {
            $li = "\t<li>";
            $li .= '<a href="'.$node->slug.'">'.$node->name.'</a>';
            if (isset($node->children)) { $li .= parseNodes($node->children); }
            $li .= "</li>\n";
            return $li;
    }


    $json = '[{"slug":"index.php","name":"Home"},{"slug":"aboutus","name":"About us","children":[{"slug":"eims","name":"Eims"},{"slug":"vision","name":"Vision"}]},{"slug":"trash","name":"Trash","children":[{"slug":"ways","name":"Ways"},{"slug":"requ","name":"Requ"}]}]';

    $nodes = json_decode($json);

    $html = parseNodes($nodes);

    echo $html;
?>

PHP：我如何将JSON放到菜单中

1 个答案: