Question

我正在将节点和链接导出到这样的json对象，以便稍后将其写入磁盘。这是我加载它时的数据：

{
"nodes": [
    {
        "id": "a1",
        "type": "activity",
        "loaded": true,
        "style": {"label": "Testing ZoomCharts"}
    },
    {
        "id": "o1",
        "type": "organization",
        "loaded": true,
        "style": {"label": "Ophileon"}
    }
],
"links": [{
    "id": "l1",
    "from": "o1",
    "to": "a1",
    "style": {"label": "Executes"}
}
}

当我从当前图表中获取节点和链接时，就像这样

    function exportNodes(){
      var nodesandlinks = {"nodes":[],"links":[]};
      nodesandlinks.nodes.push(chart._scene.data.nodes);
      nodesandlinks.links.push(chart._scene.data.links);
      alert(JSON.stringify(nodesandlinks));
    }

它返回一个我需要再次处理的结果，以便重新加载它，因为它将每个节点作为属性。

{
"nodes": [{
    "a1": {
        "id": "a1",
        "type": "activity",
        "loaded": true,
        "style": {"label": "Testing ZoomCharts"}
    },
    "o1": {
        "id": "o1",
        "type": "organization",
        "loaded": true,
        "style": {"label": "Ophileon"}
    }
}],
"links": [{"l1": {
    "id": "l1",
    "from": "o1",
    "to": "a1",
    "style": {"label": "Executes"}
}}]
}

还有另一种检索节点的方法吗？我已经尝试了这个：

chart.data.nodes
TypeError: Cannot read property 'nodes' of undefined

chart.nodes
function (){return this._impl.scene.nodes()}

Answer 1

现在有一个专用的API函数可以从网络图表中导出数据：

var exportedData = chart.exportData(visibleOnly);
alert(JSON.stringify(exportedData));

请参阅https://zoomcharts.com/developers/en/net-chart/api-reference/api/#NetChart.exportData

Answer 2

怎么样：

  function exportNodes(){
      var nodesandlinks = {"nodes":[],"links":[]};
      for (var n in chart._scene.data.nodes){
          nodesandlinks.nodes.push(chart._scene.data.nodes[n]);
      }
      for (var l in chart._scene.data.links){
          nodesandlinks.links.push(chart._scene.data.links[l]);
      }
      alert(JSON.stringify(nodesandlinks));
  }

从缩放图导出节点和链接

2 个答案: