javascript - Sharepoint 快速启动导航中的子链接

Question

https://msdn.microsoft.com/en-us/library/office/jj247080.aspx

基于此站点上的示例，用于获取快速午餐 url 和快速午餐标题

我们有这样的东西

while (nodeEnumerator.moveNext()) {

            var node = nodeEnumerator.get_current();
            nodeInfo += '{"title":"' + node.get_title() + '",' + '"link":"' + node.get_url() + '"},'; 

               }

但是如果这些导航 url 中的任何一个有任何 child ，我不知道如何得到它

那怎么得到呢？

Answer 1

使用SP.NavigationNode.children property获取导航节点的子节点集合。

Note: SP.NavigationNode.children property needs to be requested explicitly in query, this is why in the below example it is specified via Include expression: ctx.load(quickLaunchNodes,'Include(Title,Url,Children)');

示例

var ctx = SP.ClientContext.get_current();
var web = ctx.get_web();
var quickLaunchNodes = web.get_navigation().get_quickLaunch();        
ctx.load(quickLaunchNodes,'Include(Title,Url,Children)');
ctx.executeQueryAsync(function() {
        printNodesInfo(quickLaunchNodes);
    },
    function(sender, args) {
        console.log('Request failed. ' + args.get_message() + '\n' + args.get_stackTrace());
    });

function printNodesInfo(nodes){
    nodes.get_data().forEach(function(node){
        var childNodes = node.get_children();
        console.log(String.format('{0} child nodes:',node.get_title()));
        childNodes.get_data().forEach(function(childNode){
            console.log(String.format('Title: {0} Url: {1}',childNode.get_title(),childNode.get_url())); 
        });
    }); 
}