Question

我使用D3绘制气泡图，使用json格式的数据绘制气泡图我的json格式如下：{"children":[ {"name":"1","size":0.5}, {"name":"3","size":0.3636}, {"name":"4","size":0.5652}, {"name":"5","size":19.5556}, {"name":"6","size":1.037}, {"name":"7","size":0.7048}, {"name":"8","size":2.2593}, {"name":"9","size":13.7407}, {"name":"10","size":4.2222}, {"name":"11","size":2.6667}, {"name":"12","size":0.7037}, {"name":"13","size":20.6296}, {"name":"14","size":17.7037}, {"name":"15","size":2.1481}, {"name":"16","size":3.4815}, {"name":"17","size":0.1852}, {"name":"18","size":0.087}, {"name":"19","size":15.68} ]}

我正在访问这个json：

d3.json("flare.json", function(error, root) {
 var node = svg.selectAll(".node")
  .data(bubble.nodes(classes(root))
  .filter(function(d) { return !d.children; }))
.enter().append("g")
  .attr("class", "node")
  .attr("transform", function(d) { return "translate(" + d.x + "," + d.y + ")"; });

现在我想从可靠的访问 var datasource = {"children":[ {"name":"1","size":0.5}, {"name":"3","size":0.3636}, {"name":"4","size":0.5652}, {"name":"5","size":19.5556}, {"name":"6","size":1.037}, {"name":"7","size":0.7048}, {"name":"8","size":2.2593}, {"name":"9","size":13.7407}, {"name":"10","size":4.2222}, {"name":"11","size":2.6667}, {"name":"12","size":0.7037}, {"name":"13","size":20.6296}, {"name":"14","size":17.7037}, {"name":"15","size":2.1481}, {"name":"16","size":3.4815}, {"name":"17","size":0.1852}, {"name":"18","size":0.087}, {"name":"19","size":15.68} ]} 我试过像：

JSON.parse("dataSource", function(error, root) {
var node = svg.selectAll(".node")
  .data(bubble.nodes(classes(root))
  .filter(function(d) { return !d.children; }))
.enter().append("g")
  .attr("class", "node")
  .attr("transform", function(d) { return "translate(" + d.x + "," + d.y + ")"; });

但无法得到我出错的结果。请回复我。

由于 prashansa

Answer 1

datasource是变量而不是字符串

所以试试这个

JSON.parse(dataSource, function(error, root) {

EDITED

var dataSource = '{"children":...'; //JSON.parse needed
//so your code would be
JSON.parse(dataSource, function(error, root) {

var dataSource = {"children":...}; //JSON.parse not needed
//so your code would be
var node = svg.selectAll(".node")
            .data(bubble.nodes(classes(dataSource))
            .filter(function(d) { return !d.children; }))

如何从d3图表中的变量中取出json

1 个答案: