Question

我被困在一个非常简单的d.3问题......我有父子关系的节点。我正在尝试创建一个场景，我将单击除父节点之外的所有节点

我在想这样的事情呢？

.on(click, function(node){
  if(!= node.parent){
    window.location = d.url;
  }
})

显然不起作用。这是什么语法？

这就是返回的json的样子。这将给出一个想法，我如何看到父母是什么，孩子是什么

functiongetNodes(){
    varinNodes={
        "name": "Test App",
        "dept": "NYC",
        "children": [
            {
                "name": "HPD Data Feeds",
                "dept": "Third Party",
                "category": "API",
                "size": 15,
                "url": "http://nycpdev.localhost:8082/api/hpd-data-feeds"
            },
            {
                "name": "DOT Data Feeds",
                "dept": "Third Party",
                "category": "API",
                "size": 15,
                "url": "http://nycpdev.localhost:8082/api/dot-data-feeds"
            },
            {
                "name": "HPD Data Feeds",
                "dept": "Third Party",
                "category": "App",
                "size": 15,
                "url": "http://nycpdev.localhost:8082/api/hpd-data-feeds"
            }
        ],
        "size": 20,
        "url": "app/113"
    };returninNodes;
}

Answer 1

您可以直接从JSON获取节点是否有子节点：

.on(click, function(node){
  if(!node.children){
    window.location = d.url;
  }
})

单击d3.js中不是父节点的节点

1 个答案: