Question

我的要求是在JCR中迭代8000多个节点，并使用PageManager API为每个节点用Java创建一个Page对象。

首先，我使用PageManager获取页面的标题，如下所示。

public String currentPageTitle(String pagePath) {
    Page page=null;
    ResourceResolver resourceResolver=null;
    PageManager pageManager=null;
    try {
        if (pagePath != null) {
            resourceResolver = resourceResolverFactory.getAdministrativeResourceResolver(null);
            pageManager = resourceResolver.adaptTo(PageManager.class);
            **page = pageManager.getContainingPage(resourceResolver.getResource(pagePath));**
            LOGGER.error("Page $$$$"+page);
            if (page == null) {
                throw new IllegalArgumentException("Page does not exist: " + pagePath);
            }
        }
    } catch (LoginException e) {
        LOGGER.error("Login Exception");
        e.printStackTrace();
    }
    return page.getTitle();
}

这里我将页面对象设为null，当我尝试传递Geometrixx页面URL以获取其标题时，它正在抛出“页面不存在：/content/geometrixx/fr.html”。

Answer 1

删除扩展名（.html）和Execute.It将正常工作。对于超过80000页的迭代，使用递归函数。

public String currentPageTitle(String pagePath) {
    Page page=null;
    ResourceResolver resourceResolver=null;
    PageManager pageManager=null;
    try {
        if (pagePath != null) {
            resourceResolver = resourceResolverFactory.getAdministrativeResourceResolver(null);
            pageManager = resourceResolver.adaptTo(PageManager.class);
            page = pageManager.getContainingPage(resourceResolver.getResource(pagePath));
            LOGGER.error("Page $$$$"+page);
            if (page == null) {
                throw new IllegalArgumentException("Page does not exist: " + pagePath);
            }else{
                buildLinkAndChildren_loop(page);
            }
        }
    } catch (LoginException e) {
        LOGGER.error("Login Exception");
        e.printStackTrace();
    }
    return page.getTitle();
}
public void buildLinkAndChildren_loop(Page page) {
    if (page != null) {
        Iterator<Page> children = page.listChildren();
        while (children.hasNext()) {
            Page child = children.next();
            buildLinkAndChildren_loop(child);

        }
    }
}

在AEM6 / CQ5中使用PageManager API

1 个答案: