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javascript - 如何在D3.js中实现Tree的双击?

转载 作者:行者123 更新时间:2023-12-02 15:51:54 25 4
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我是 D3.js 新手。

用于绘制树的JS:

var json = 
{
"name": "Base",
"children": [
{
"name": "Type A",
"children": [
{
"name": "Section 1",
"children": [
{"name": "Child 1"},
{"name": "Child 2"},
{"name": "Child 3"}
]
},
{
"name": "Section 2",
"children": [
{"name": "Child 1"},
{"name": "Child 2"},
{"name": "Child 3"}
]
}
]
},
{
"name": "Type B",
"children": [
{
"name": "Section 1",
"children": [
{"name": "Child 1"},
{"name": "Child 2"},
{"name": "Child 3"}
]
},
{
"name": "Section 2",
"children": [
{"name": "Child 1"},
{"name": "Child 2"},
{"name": "Child 3"}
]
}
]
}
]
};

var width = 700;
var height = 650;
var maxLabel = 150;
var duration = 500;
var radius = 5;

var i = 0;
var j = 0;
var root;
var columnAttribute ;

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

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

var svg = d3.select("body").append("svg")
.attr("width", width)
.attr("height", height)
.append("g")
.attr("transform", "translate(" + maxLabel + ",0)")
;

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

root.children.forEach(collapse);

function update(source)
{
// Compute the new tree layout.
var nodes = tree.nodes(root).reverse();
var links = tree.links(nodes);

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

// 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(" + source.y0 + "," + source.x0 + ")"; });

nodeEnter.append("circle")
.attr("r", 0)
.style("fill", function(d){
return d._children ? "lightsteelblue" : "white";
});

nodeEnter.append("text")
.attr("x", function(d){
var spacing = computeRadius(d) + 5;
return d.children || d._children ? -spacing : spacing;
})
.attr("dy", "3")
.attr("text-anchor", function(d){ return d.children || d._children ? "end" : "start"; })
.text(function(d){ return d.name; })
.style("fill-opacity", 0);

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

nodeUpdate.select("circle")
.attr("r", function(d){ return computeRadius(d); })
.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(" + source.y + "," + source.x + ")"; })
.remove();

nodeExit.select("circle").attr("r", 0);
nodeExit.select("text").style("fill-opacity", 0);

// 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: source.x0, y: source.y0};
return diagonal({source: o, target: o});
});

// 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: 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;
});
}

function computeRadius(d)
{
if(d.children || d._children) return radius + (radius * nbEndNodes(d) / 10);
else return radius;
}

function nbEndNodes(n)
{
nb = 0;
if(n.children){
n.children.forEach(function(c){
nb += nbEndNodes(c);
});
}
else if(n._children){
n._children.forEach(function(c){
nb += nbEndNodes(c);
});
}
else nb++;

return nb;
}

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

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

update(root);

我想在双击 columnAttribute 变量中的节点时保存节点名称。

例如,单击名为 Child 1 的节点时,它保存在 columnAttribute : ["Child 1"] 中。同样,单击名为 Child 2 的节点,它保存在 columnAttribute 中:["Child 1","Child2"]

JSFiddle:http://jsfiddle.net/MetalMonkey/JnNwu/

最佳答案

您可以在新创建的节点上使用双击事件(称为dblclick):

var nodeEnter = node.enter()
.append("g")
.attr("class", "node")
.attr("transform", function(d){ return "translate(" + source.y0 + "," + source.x0 + ")"; })
.on("click", click)
.on("dblclick", dblClick)

dblClick函数中,只需将名称或任何您想要的内容添加到columnAttribute:

var columnAttribute = [];
function dblClick(d) {
columnAttribute.push(d.name);
console.log(columnAttribute);
}

查看此fiddle

关于javascript - 如何在D3.js中实现Tree的双击?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/31789449/

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