Question

我希望在点击它时更改节点的颜色。它应该在＆＃34;点击＆＃34;功能。

然而，在点击功能中，＆＃34; d＆＃34;是节点元素，但要更改节点颜色，我必须选择＆＃34; circle＆＃34; element，使用nodeEnter.append附加到节点（＆＃34; circle＆＃34;）。我怎样才能实现它？

html文件：

<!DOCTYPE html>
<meta charset="utf-8">
<style>
.node circle {
fill: #fff;
stroke: steelblue;
        stroke-width: 3px;
}

.node text {
font: 30px sans-serif;
}

.link {
fill: none;
stroke: #ccc;
        stroke-width: 7px;

}
</style>
<body>
<script src="http://d3js.org/d3.v3.min.js"></script>
<script>


var margin = {top: 20, right: 20, bottom: 20, left: 20},
    width = 1200 - margin.right - margin.left,
    height = 800 - margin.top - margin.bottom;

var i = 0,
    duration = 750,
    root;

var tree = d3.layout.tree();

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

    var svg = d3.select("body").append("svg")
    .attr("width", width + margin.right + margin.left)
    .attr("height", height + margin.top + margin.bottom)
    .append("g")
    .attr("transform", "translate(" + margin.left + "," + margin.top + ")");


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

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

            tree.size([width, height]);

            update(root);
            });

function update(source1) {
    var originalConsole = console;
    var nodes = tree.nodes(root);
    var links = tree.links(nodes);

    nodes.forEach(function(d) { d.y = d.depth * 100; });

    // Update the nodes…
    var node = svg.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(" + source1.y0 + "," + source1.x0 + ")"; })
        .on("click", click);

    nodeEnter.append("circle")
        .attr("r", 1e-6)
        .style("fill", function(d) { return d._children ? "lightsteelblue" : "#fff"; });

    nodeEnter.append("text")
        .attr("x", function(d) { return d.children || d._children ? -20 : 20; })
        .attr("dy", ".35em")
        .attr("text-anchor", function(d) { return d.children || d._children ? "end" : "start"; })
        .text(function(d) { return d.name; })
        .style("fill-opacity", 1e-6);


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

    nodeUpdate.select("circle")
        .attr("r", 10)
        .style("fill", function(d) { return d._children ? "lightsteelblue" : "#fff"; });

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

    // Transition exiting nodes to the parent's new position.
    var nodeExit = node.exit().transition()
        .duration(duration)
        .attr("transform", function(d) { return "translate(" + source1.y + "," + source1.x + ")"; })
        .remove();

    nodeExit.select("circle")
        .attr("r", 1e-6);

    nodeExit.select("text")
        .style("fill-opacity", 1e-6);

    // Update the links…
    var link = svg.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: source1.x0, y: source1.y0};
                return diagonal({source: o, target: o});
                })
    .style("stroke-width", "3px")
        .style("stroke", "green");
    // Transition links to their new position.
    link.transition()
        .duration(duration)
        .attr("d", diagonal);

    // Transition exiting nodes to the parent's new position.
    link.exit().transition()
        .duration(duration)
        .attr("d", function(d) {
                var o = {x: source1.x, y: source1.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;

            });
    for(var k = 0; k < 1000; k++)
        flag = 0;
}

// Toggle children on click.
function click(d) {
    if (d.children) {
        d._children = d.children;
        d.children = null;
    } else {
        d.children = d._children;
        d._children = null;
    }
    update(d);
}


</script>

json文件：

         {"name":"VP",
          "size":"89",
         "children": [
         {"name":"VBP",
          "size":"15",
         "children":[{"name":"are", "size":"38"}]
         },
         {"name":"NP",
          "size":"83",
         "children": [
         {"name":"DT",
          "size":"29",
         "children":[{"name":"a", "size":"53"}]
         },
         {"name":"NN",
          "size":"50",
         "children":[{"name":"boy", "size":"99"}]
         }
         ]
         }
         ]
         }

Answer 1

d 是数据，而不是节点

点击功能中的

此是dom节点

d3.select（this）是仅选择该节点作为d3选择的选择

所以要在选择中获得子圆元素，你会这样做 d3.select(this).select("circle")

现在你认为你只需要添加一个style("fill", "red")或其他任何内容和完成工作，但因为这是来自Mike Bostock的常见例子，在转换更新中样式将被再次覆盖。因此，您需要做的是在数据上设置属性，然后在输入/更新/退出序列中通过绘图获取该属性。

最后，点击功能所需的只有：

d.clicked = true;

并在nodeUpdate部分执行此操作 - ＆gt;

nodeUpdate.select("circle")
        .attr("r", function(d){ return computeRadius(d); })
        .style("fill", function(d) { return d.clicked ? "red" : (d._children ? "lightsteelblue" : "#fff");});

请注意，这对取消突出显示节点没有任何作用。（您必须切换之前的d.clicked布尔值）

D3.js Tree：点击它时如何更改节点颜色？

1 个答案: