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使用 JavaScript Use-Api 我能够创建一个自定义对象并将其返回到一个 html 文件。此功能允许我创建自定义对象列表，可用于创建菜单或其他复杂的类似列表的组件。

假设我有以下内容结构：

/content
    /project
        /homepage
            /contentpage1
                /contentpage1.1
                /contentpage1.2
                /contentpage1.3 (hidden)
            /contentpage2
                /contentpage1.1 (hidden)
                /contentpage1.2 (hidden)
                /contentpage1.3 (hidden)
            /contentpage3
            /contentpage4

菜单应仅包含第一级内容页。每个菜单项都应该有带有二级内容页面的下拉列表，如果它们存在并且没有隐藏的话。我可以使用以下代码在 JavaScript 中执行此操作：

"use strict";

use(function() {

    function getMenuItems() {
        var currentPageDepth = currentPage.getDepth();
        var menuObjects = [];
        if(currentPageDepth >= 3) {
            var homePage = currentPage.getAbsoluteParent(2);
            var list = homePage.listChildren();
            while(list.hasNext()) {
                var tempPage = list.next()
                var customPageObject = createMenuItemObject(tempPage);
                menuObjects.push(customPageObject);
            }            
        }
        return menuObjects;
    }

    function createMenuItemObject(page) {
        // ...
        // looking for any other properties of page or its children
        // ...
        return {page: page, 
                visibleChildrenExists: visibleChildrenExists(page)};
    }

    function visibleChildrenExists(page) {
        var list = page.listChildren();
        var visibleChildrenExists = false;
        while(list.hasNext()) {
            var subPage = list.next();
            if(!subPage.isHideInNav()) {
                visibleChildrenExists = true;
                break;
            }
        }
        return visibleChildrenExists;
    }

    return {
        menuObjectsList: getMenuItems(),
    };
}

HTML：

<headerComponent data-sly-use.headerComponentJS="headerComponent.js" data-sly-unwrap />
<menuItems data-sly-list.menuItem="${headerComponentJS.menuObjectsList}"  data-sly-unwrap >
     <li class='${menuItem.visibleChildrenExists ? "" : "direct"}' data-sly-test="${!menuItem.page.hideInNav}">
        <a href="${menuItem.page.path}.html">${menuItem.page.title}</a>
        <ul data-sly-test="${menuItem.visibleChildrenExists}" data-sly-list.submenuItem="${menuItem.page.listChildren}">
            <li data-sly-test="${!submenuItem.hideInNav}">
                <a href="${submenuItem.path}.html">${submenuItem.title}</a>
            </li>
        </ul>
     </li>       
</menuItems>

为什么要使用 Java Use-Api？在 Resource 或 Node 等接口上操作更容易。看起来它在 JavaScript 中不能很好地工作，但我需要有可能返回具有多个属性的自定义对象。

The question is: is it even possible to do something similar using Java Use-Api? What do I have to return? I can't return a map, because it won't be possible to access its elements since it's not possible to pass a parameter to Java Use-Api method... Any suggestion?

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1 回答 1

1

It is possible to return maps using the java-use api see an example below:

Method in the Java class

//Return a map
public Map<String, String> getTestMap() {
    //TODO some coding
    Map<String,String> testMap = new HasMap<String,String>();
    testMap.put("IDA", "test value");
    testMap.put("IDB", "test value 2");

    return testMap;     
}

HTML code to access each element of the map:

<div data-sly-use.param="JavaClass">    
    <div data-sly-test.map="${param.testMap}">
        <div class="pos">
            <span class="classA">${map['IDA']}</span><br>
            <span class="classB">${map['IDB']}</span>
        </div>
    </div>
</div>
于 2015-02-06T14:05:16.810 回答