Question

我试图让树节点排成一行（路径），我无法解决。

这是一个从上到下的树，我试图翻到左到右的树，但我认为计算结束了。

任何帮助都会非常感激。

请参阅Plunker中的截图和代码。 http://plnkr.co/edit/9pxJLz?p=preview

var margin = {
        top: 20,
        right: 120,
        bottom: 20,
        left: 120
      },
      width = 960 - margin.right - margin.left,
      height = 100 - margin.top - margin.bottom;

    var i = 0,
      duration = 750,
      rectW = 120,
      rectH = 30;

    var tree = d3.layout.tree().nodeSize([height, width]);

    var diagonal = d3.svg.diagonal()
      .projection(function(d) {
        return [d.y, d.x];
      });

    var accountSvg = d3.select("body").append("svg")
      .attr("width", 1000)
      .attr("height", 1000)
      .call(zm = d3.behavior.zoom().scaleExtent([0.5, 3]).on("zoom", redraw)).on("dblclick.zoom", null)
      .append("g")
      .attr("transform", "translate(" + margin.left + "," + margin.left + ")");

    d3.json("flare.json", function(error, flare) {
      if (error) throw error;

      root = flare;
      root.x0 = height / 2;
      root.y0 = 0;

      function collapse(d) {
        if (d.children) {
          d._children = d.children;
          d._children.forEach(collapse);
          d.children = null;
        }
      }

      root.children.forEach(collapse);
      update(root);
    });

    //necessary so that zoom knows where to zoom and unzoom from
    zm.translate([250, 20]);

    d3.select("body").style("height", "455");

    function update(source) {

      // Compute the new tree layout.
      var nodes = tree.nodes(root).reverse(),
        links = tree.links(nodes);

      // Normalize for fixed-depth.
      nodes.forEach(function(d) {
        d.y = d.depth * 180;
      });

      // Update the nodes…
      var node = accountSvg.selectAll("g.node")
        .data(nodes, function(d) {
          return d.id || (d.id = ++i);
        });

      // Enter any new nodes at the parent's previous position.
      var nodeEnter = node.enter().append("g")
        .attr("class", "node")
        .attr("transform", function(d) {
          return "translate(" + source.y0 + "," + source.x0 + ")";
        })
        .on("click", click);

      nodeEnter.append("rect")
        .attr("width", rectW)
        .attr("height", rectH)
        .attr("stroke", "black")
        .attr("stroke-width", 1)
        .style("fill", function(d) {
          return d._children ? "lightsteelblue" : "#fff";
        });

      nodeEnter.append("text")
        .attr("x", rectW / 2)
        .attr("y", rectH / 2)
        .attr("dy", ".35em")
        .attr("text-anchor", "middle")
        .text(function(d) {
          return d.name;
        });

      // Transition nodes to their new position.
      var nodeUpdate = node.transition()
        .duration(duration)
        .attr("transform", function(d) {
          return "translate(" + d.y + "," + d.x + ")";
        });

      nodeUpdate.select("rect")
        .attr("width", rectW)
        .attr("height", rectH)
        .attr("stroke", "black")
        .attr("stroke-width", 1)
        .style("fill", function(d) {
          return d._children ? "lightsteelblue" : "#fff";
        });

      nodeUpdate.select("text")
        .style("fill-opacity", 1)
        .style("fill", '#404080');

      // Transition exiting nodes to the parent's new position.
      var nodeExit = node.exit().transition()
        .duration(duration)
        .attr("transform", function(d) {
          return "translate(" + source.y + "," + source.x + ")";
        })
        .remove();

      nodeExit.select("rect")
        .attr("width", rectW)
        .attr("height", rectH)
        .attr("stroke", "black")
        .attr("stroke-width", 1);

      nodeExit.select("text");

      // Update the links…
      var link = accountSvg.selectAll("path.link")
        .data(links, function(d) {
          return d.target.id;
        });

      // Enter any new links at the parent's previous position.
      link.enter().insert("path", "g")
        .attr("class", "link")
        .attr("d", function(d) {
          var o = {
            x: source.x0,
            y: source.y0
          };
          return diagonal({
            source: o,
            target: o
          });
        });

      // Transition links to their new position.
      link.transition()
        .duration(duration)
        .attr("d", function(d) {
          var s = {
            x: d.source.x0 + -15,
            y: d.source.y0 + 120
          }
          var t = {
            x: d.target.x + 15,
            y: d.target.y + 120
          }
          return diagonal({
            source: s,
            target: t
          })
        });
      //.attr("d", diagonal);

      // Transition exiting nodes to the parent's new position.
      link.exit().transition()
        .duration(duration)
        .attr("d", function(d) {
          var o = {
            x: source.x,
            y: source.y
          };
          return diagonal({
            source: o,
            target: o
          });
        })
        .remove();

      // Stash the old positions for transition.
      nodes.forEach(function(d) {
        d.x0 = d.x;
        d.y0 = d.y;
      });
    }

    var doubleClickTime = 0;
    var threshold = 200;
    // Toggle children on click.
    function click(d) {
      var t0 = new Date();
      if (t0 - doubleClickTime > threshold) {
        setTimeout(function() {
          if (t0 - doubleClickTime > threshold) {

            if (d.children) {
              d._children = d.children;
              d.children = null;
            } else {
              d.children = d._children;
              d._children = null;
            }
            update(d);

          }
        }, threshold);
      }
    }

    // Redraw for zoom
    function redraw() {
      //console.log("here", d3.event.translate, d3.event.scale);
      accountSvg.attr("transform",
        "translate(" + d3.event.translate + ")" + " scale(" + d3.event.scale + ")");
    }

Answer 1

您正在翻译所有链接：

.attr("transform", function (d) {
    return "translate(" + rectW + "," + rectH / 2 + ")";
})

不要那样做。而不是那样，改变对角线：

link.transition()
    .duration(duration)
    .attr("d", function(d) {
        var s = {
            y: d.source.y + rectW,
            x: d.source.x + rectH / 2
        };
        var t = {
            x: d.target.x + rectH / 2,
            y: d.target.y
        };
        return diagonal({
            source: s,
            target: t
        })
    });

以下是更新的plunker：http://plnkr.co/edit/LnuoQY7R0tDWR4EnW1Rg?p=preview

d3.v3具有可折叠框的水平树结构

1 个答案: