如何仅删除两个节点之间的路径而不删除节点

Question

这是我的图表

const data = {
  "name": "tree",
  "id": 0,
  "children": [{
    "name": "A",
    "id": 1,
    "children": [{
        "name": "B",
        "id": 2,

        "children": [{
          "name": "C",
          "id": 3
        }, ]
      },
      {
        "name": "H",
        "id": 4,

        "children": [{
            "name": "D",
            "id": 5
          },
          {
            "name": "E",
            "id": 6
          },
          {
            "name": "F",
            "id": 7
          },
        ]
      },
    ]
  }]
}

const width = 928;

// Compute the tree height; this approach will allow the height of the
// SVG to scale according to the breadth (width) of the tree layout.
const root = d3.hierarchy(data);
const dx = 10;
const dy = width / (root.height + 1);

// Create a tree layout.
const tree = d3.tree().nodeSize([dx, dy]);

// Sort the tree and apply the layout.
root.sort((a, b) => d3.ascending(a.data.name, b.data.name));
tree(root);

// Compute the extent of the tree. Note that x and y are swapped here
// because in the tree layout, x is the breadth, but when displayed, the
// tree extends right rather than down.
let x0 = Infinity;
let x1 = -x0;
root.each(d => {
  if (d.x > x1) x1 = d.x;
  if (d.x < x0) x0 = d.x;
});

// Compute the adjusted height of the tree.
const height = x1 - x0 + dx * 2;

const svg = d3.select("svg")
  .attr("width", width)
  .attr("height", height)
  .attr("viewBox", [-dy / 3, x0 - dx, width, height])
  .attr("style", "max-width: 100%; height: auto; font: 10px sans-serif;");

const link = svg.append("g")
  .attr("fill", "none")
  .attr("stroke", "#555")
  .attr("stroke-opacity", 0.4)
  .attr("stroke-width", 1.5)

link
  .selectAll()
  .data(root.links())
  .join("path")
  .attr("d", d3.linkHorizontal()
    .x(d => d.y)
    .y(d => d.x));

const node = svg.append("g")
  .attr("stroke-linejoin", "round")
  .attr("stroke-width", 3)
  .selectAll()
  .data(root.descendants())
  .join("g")
  .attr("transform", d => `translate(${d.y},${d.x})`)

node.append("circle")
  .attr("fill", d => d.children ? "#555" : "#999")
  .attr("r", 5.5)

node.append("text")
  .attr("dy", "0.31em")
  .attr("x", d => d.children ? -6 : 6)
  .attr("text-anchor", d => d.children ? "end" : "start")
  .text(d => d.data.name)
  .clone(true).lower()
  .attr("stroke", "white");

d3.selectAll("circle").on("contextmenu", (event) => {
  this.selectedNode = event.srcElement.__data__.data
});

const connectNodes = (t, f) => {
  let to = null,
    fr = null;
  node.each(d => {
    if (d.data.name === t) to = d;
    if (d.data.name === f) fr = d;
  });
  if (to && fr) {
    link.append("path")
      .attr("d", "M" + to.y + "," + to.x + "L" + fr.y + "," + fr.x)
      .attr("fill", "none")
      .attr("stroke", "red");
  }
};
connectNodes("B", "D");

<!doctype html>

<html>

<head>
  <script src="https://cdnjs.cloudflare.com/ajax/libs/d3/7.8.5/d3.js"></script>

</head>

<body>
  <svg></svg>

</body>

</html>

我想删除 node H 和 node F 之间的路径而不删除 F 节点

结果如上图所示。

我需要完成node id

function removePath(from_node_id:number, to_node_id:number){
  // 
}

我也有这个功能，但这不是我想要的，因为它接受节点名称

const removeConnection = (t, f) => {
  d3.select("#" + t + "-" + f).remove();
}

removeConnection("H", "F");

Answer 1

如果 ypur 函数没问题，并且你只想要 name 和 id 之间的关系，只需创建一个像这样的 dico （递归函数）：

let dico = {};
getDico(data);
console.log(Object.keys(dico), dico["B"]);
function getDico(s){

    if(Array.isArray(s)){
    s.forEach( d => {
        dico[d.name] = d.id;
      if(d.hasOwnProperty("children"))
                getDico(d.children);
      });
  } else { // its not an array
    dico[s.name] =s.id;
    if(s.hasOwnProperty("children"))
        getDico(s.children); 
  }
}

如何仅删除两个节点之间的路径而不删除节点

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如何仅删除两个节点之间的路径而不删除节点

问题描述 投票：0回答：1

1个回答

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问题描述投票：0回答：1