在SAPUI5的SplitApp中从一个列表导航到另一个列表

<mvc:View xmlns:mvc="sap.ui.core.mvc" xmlns="sap.m">
    <SplitApp id="mySplitApp">
        <!-- 左侧列表 -->
        <nav>
            <List id="list1" items="{/items}" selectionChange="onListItemPress">
                <StandardListItem title="{title}" />
            </List>
        </nav>
        <!-- 右侧列表 -->
        <detailPages>
            <Page id="page1">
                <List id="list2" items="{/subItems}">
                    <StandardListItem title="{title}" />
                </List>
            </Page>
        </detailPages>
    </SplitApp>
</mvc:View>

sap.ui.define([
    "sap/ui/core/mvc/Controller",
    "sap/ui/core/routing/History"
], function(Controller, History) {
    "use strict";

    return Controller.extend("myController", {
        onListItemPress: function(oEvent) {
            var oItem = oEvent.getParameter("listItem");
            var oContext = oItem.getBindingContext();
            var sPath = oContext.getPath();

            // 获取SplitApp控件
            var oSplitApp = this.getView().byId("mySplitApp");

            // 获取右侧列表页面
            var oPage = this.getView().byId("page1");

            // 设置右侧列表的上下文路径
            oPage.bindElement({
                path: sPath,
                model: "modelName"
            });

            // 导航到右侧列表页面
            oSplitApp.toDetail(oPage.getId());
        }
    });
});