java - 递归构建 <ul> 列表

Question

所以，我有这个网站结构

Page 1
  Page 1.1
Page 2
  Page 2.1
    Page 2.1.1
  Page 2.2
Page 3
  Page 3.1
  Page 3.2
Page 4

我想构建<ul>使用递归函数列表。我的函数如下所示

public String getMenu(Page rootPage, boolean base){
    final Logger log = LoggerFactory.getLogger(this.getClass());
    Iterator<Page> subPages = rootPage.listChildren();
    StringBuilder output = new StringBuilder("<ul");
        output.append(" id=\"drop-menu\"");
    output.append(" class=\"popup-menu\">");
    if(!base){
        output.append("<li><a href=\"").append(rootPage.getPath()).append(".html\" class=\"showSubPage\" rel=\"").append(rootPage.getPath()).append("\">");
        String title = rootPage.getPageTitle() == null ? rootPage.getTitle() : rootPage.getPageTitle();
        output.append(title);
        output.append("</a>");
        output.append("</li>");
        output.append("</ul>");
    }
    while(subPages.hasNext()){
        output.append("<ul>");
        log.info("som subpages here!");
        Page curPage = subPages.next();
        output.append("<li><a href=\"").append(curPage.getPath()).append(".html\" class=\"showSubPage\" rel=\"").append(curPage.getPath()).append("\">");
        String title = curPage.getPageTitle() == null ? curPage.getTitle() : curPage.getPageTitle();
        output.append(title);
        output.append("</a>");
        Iterator<Page> subSub = curPage.listChildren();
        int tmpCtr = 0;
        while(subSub.hasNext()){
            tmpCtr++;
            output.append(getMenu(subSub.next(), false));
        }
        output.append("</li>");
        output.append("</ul>");
    }
    return output.toString();
}

输出看起来像这样

<ul id="drop-menu" class="popup-menu">
    <ul>
        <li><a href="/menu-hier/afsafa.html" class="showSubPage" >Page 1</a>
            <ul id="drop-menu" class="popup-menu">
                <li><a href="/menu-hier/afsafa/sdgsdgdsf.html" class="showSubPage" rel="/menu-hier/afsafa/sdgsdgdsf">Page 1.1</a>
                </li>
            </ul>
        </li>
    </ul>
    <ul>
        <li><a href="/menu-hier/fdsafdsafas.html" class="showSubPage" rel="/menu-hier/fdsafdsafas">Page 2</a>
            <ul id="drop-menu" class="popup-menu">
                <li><a href="/menu-hier/fdsafdsafas/sdfgdsgsdg.html" class="showSubPage" rel="/menu-hier/fdsafdsafas/sdfgdsgsdg">Page 2.1</a>
                </li>
            </ul>
            <ul>
                <li><a href="/menu-hier/fdsafdsafas/sdfgdsgsdg/dsgsdgdsgfd.html" class="showSubPage" rel="/menu-hier/fdsafdsafas/sdfgdsgsdg/dsgsdgdsgfd">Page 2.1.1</a>
                </li>
            </ul>
            <ul id="drop-menu" class="popup-menu">
                <li><a href="/menu-hier/fdsafdsafas/fsdgdsfgsdgfs.html" class="showSubPage" rel="/menu-hier/fdsafdsafas/fsdgdsfgsdgfs">Page 2.2</a>
                </li>
            </ul>
        </li>
    </ul>
    <ul>
        <li><a href="/menu-hier/fgdsgfsdgdsfgsdg.html" class="showSubPage" rel="/menu-hier/fgdsgfsdgdsfgsdg">Page 3</a>
            <ul id="drop-menu" class="popup-menu">
                <li><a href="/menu-hier/fgdsgfsdgdsfgsdg/fsafsdafas.html" class="showSubPage" rel="/menu-hier/fgdsgfsdgdsfgsdg/fsafsdafas">Page 3.1</a>
                </li>
            </ul>
            <ul id="drop-menu" class="popup-menu">
                <li><a href="/menu-hier/fgdsgfsdgdsfgsdg/yxvcxyvvyxcv.html" class="showSubPage" rel="/menu-hier/fgdsgfsdgdsfgsdg/yxvcxyvvyxcv">Page 3.2</a>
                </li>
            </ul>
            <ul id="drop-menu" class="popup-menu">
                <li><a href="/menu-hier/fgdsgfsdgdsfgsdg/yxvcxyv.html" class="showSubPage" rel="/menu-hier/fgdsgfsdgdsfgsdg/yxvcxyv">Page 3.3</a>
                </li>
            </ul>
        </li>
    </ul>
    <ul>
        <li><a href="/menu-hier/dsfgsdfgdsfg.html" class="showSubPage" rel="/menu-hier/dsfgsdfgdsfg">Page 4</a>
        </li>
    </ul>

所以问题是，3 级页面放置不正确。例如，页面 2.1.1 不在页面 2.1 部分下。

感谢您的帮助!

Answer 1

不确定您希望 HTML 是什么样子，但是:

1) 从您的代码中，<ul>标签在子页面中插入两次(一次在 while 循环中，然后再次在递归调用 getMenu() 中)

2)我认为，您错过了 <li> <ul> 下面的标签标签。

3)你的代码看起来相当冗余和复杂，难道不能像这样更容易地完成(未经测试):

public String getMenu(Page page, boolean isRoot) {
    StringBuilder output = new StringBuilder();
    if (isRoot) {
        output.append("<ul id=\"drop-menu\"");
        output.append(" class=\"popup-menu\">");
        }
    else {
        output.append("<ul>");
        }
    output.append("<li><a href=\"")
        .append(page.getPath())
        .append(".html\" class=\"showSubPage\" rel=\"")
        .append(rootPage.getPath()).append("\">");
    String title = page.getPageTitle() == null ? page.getTitle() : page.getPageTitle();
    output.append(title);
    output.append("</a>");

    Iterator<Page> subPages = page.listChildren();
    while(subPages.hasNext()){
        output.append(getMenu(subPages.next(), false));
    }
    output.append("</li>");
    output.append("</ul>");
    return output.toString();
}