问题是我想在树中添加不同节点颜色的每个节点，我尝试了多个解决方案，但没有结果。下面是我的数据，如何更改我的数据以将其添加到具有不同颜色的树。 这是我的Tree
var treeData = {"name" : "HTTP-GET", "children" : [
{"name" : " wah.comsats.edu.pk/ ", "children" : [
{"name" : " wah.comsats.edu.pk/AboutCIIT/Introduction.aspx","children" :[
{"name" : " wah.comsats.edu.pk/AboutCIIT/Introduction.aspx " },
{"name" : " wah.comsats.edu.pk/AboutCIIT/Introduction.aspx " },
{"name" : " Embedded Stream ", "children" : [
{"name" : " wah.comsats.edu.pk/favicon.ico " },
{"name" : " wah.comsats.edu.pk/favicon.ico " }]}]},
{"name" : " wah.comsats.edu.pk/AboutCIIT/Introduction.aspx " },
{"name" : " wah.comsats.edu.pk/AboutCIIT/Introduction.aspx " },
{"name" : " Embbeded Stream ", "children" : [
{"name" : " wah.comsats.edu.pk/favicon.ico " },
{"name" : " wah.comsats.edu.pk/favicon.ico " }]}]},
{"name" : " cuonline.ciitwah.edu.pk/ "}
]};
以下是我可视化树的代码。我试图改变功能，但它停止了树的可视化。
<!DOCTYPE html>
<meta charset="UTF-8">
<style>
.node circle {
fill: #fff;
stroke: steelblue;
stroke-width: 1px;
}
.node text {
font: 12px sans-serif;
text-color:white;
}
.link {
fill: none;
stroke: #ccc;
stroke-width: 3px;
}
body{
background-image: url("gray.png");
}
</style>
<body>
<script type="text/javascript" src="lib/d3.min.js"></script>
<script>
var treeData = {"name" : "HTTP-GET", "children" : [
{"name" : " http://wah.comsats.edu.pk/ ", "children" : [
{"name" : " http://fonts.googleapis.com/css?family=Oswald " },
{"name" : " http://wah.comsats.edu.pk/img/acad.jpg " },
{"name" : " http://wah.comsats.edu.pk/js/responsiveCarousel.min.js " },
{"name" : " http://wah.comsats.edu.pk/slides/wahm.jpg " },
{"name" : " http://wah.comsats.edu.pk/slides/BITA17.jpg " },
{"name" : " http://wah.comsats.edu.pk/slides/alumni.jpg " },
{"name" : " http://wah.comsats.edu.pk/img/ann.jpg " },
{"name" : " http://wah.comsats.edu.pk/img/ann.jpg " }]},
{"name" : " https://www.google.com.pk/search?q=nsut&oq=nsut&aqs=chrome..69i57j69i60l4j69i59.613j0j7&sourceid=chrome&ie=UTF-8 "},
{"name" : " https://www.google.com.pk/url?sa=t&source=web&rct=j&url=http://www.nsut.com/&ved=0ahUKEwiq-q3goqbaAhXSxqQKHfLhBDIQFggmMAA "},
{"name" : " https://www.google.com.pk/search?q=nust&oq=nust&aqs=chrome..69i57j0l5.814j0j7&sourceid=chrome&ie=UTF-8 "},
{"name" : " https://pbs.twimg.com/profile_images/734623083781345280/Kb-33Gf3_normal.jpg "},
{"name" : " https://pbs.twimg.com/media/DaGZdo6V4AAJ6PW?format=jpg&name=medium "},
{"name" : " https://www.google.com.pk/search?q=patent&oq=patent&aqs=chrome..69i57j0l5.6101j0j7&sourceid=chrome&ie=UTF-8 "},
{"name" : " https://syndication.twitter.com/i/jot "},
{"name" : " https://platform.twitter.com/jot.html "},
]};
var margin = {top: 20, right: 90, bottom: 30, left: 90},
width = 1960 - margin.left - margin.right,
height = 500 - margin.top - margin.bottom;
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 + ")");
var i = 0,
duration = 750,
root;
var treemap = d3.tree().size([height, width]);
root = d3.hierarchy(treeData, function(d) { return d.children; });
root.x0 = height / 2;
root.y0 = 0;
root.children.forEach(collapse);
update(root);
function collapse(d) {
if(d.children) {
d._children = d.children
d._children.forEach(collapse)
d.children = null
}
}
function update(source) {
var treeData = treemap(root);
var nodes = treeData.descendants(),
links = treeData.descendants().slice(1);
nodes.forEach(function(d){ d.y = d.depth * 180});
var node = svg.selectAll('g.node')
.data(nodes, function(d) {return d.id || (d.id = ++i); });
var nodeEnter = node.enter().append('g')
.attr('class', 'node')
.attr("transform", function(d) {
return "translate(" + source.y0 + ", " + source.x0 + ")";
})
.on('click', click);
nodeEnter.append('circle')
.attr('class', 'node')
.attr('r', 1e-6)
.style("fill", function(d,i) {
return d._childern ? "lightsteelblue" : "#fff";
});
nodeEnter.append('text')
.attr("dy", ".35em")
.attr("x", function(d) {
return d.children || d._children ? -13 : 13;
})
.attr("text-anchor", function(d) {
return d.children || d._children ? "end" : "start";
})
.text(function(d) { return d.data.name; });
var nodeUpdate = nodeEnter.merge(node);
nodeUpdate.transition()
.duration(duration)
.attr("transform", function(d) {
return "translate(" + d.y + "," + d.x + ")";
});
nodeUpdate.select('circle.node')
.attr('r', 10)
.style("fill", function(d) {
return d._children ? "lightsteelblue" : "#fff";
})
.attr('cursor', 'pointer');
var nodeExit = node.exit().transition()
.duration(duration)
.attr("transform", function(d) {
return "translate(" + source.y + "," + source.x + ")";
})
.remove();
nodeExit.select('circle')
.attr('r', 1e-6);
nodeExit.select('text')
.style('fill-opacity', 1e-6);
var link = svg.selectAll('path.link')
.data(links, function(d) { return d.id; });
var linkEnter = link.enter().insert('path', "g")
.attr("class", "link")
.attr('d', function(d){
var o = {x: source.x0, y: source.y0}
return diagonal(o, o)
});
var linkUpdate = linkEnter.merge(link);
linkUpdate.transition()
.duration(duration)
.attr('d', function(d){ return diagonal(d, d.parent) });
var linkExit = link.exit().transition()
.duration(duration)
.attr('d', function(d) {
var o = {x: source.x, y: source.y}
return diagonal(o, o)
})
.remove();
nodes.forEach(function(d){
d.x0 = d.x;
d.y0 = d.y;
});
function diagonal(s, d) {
path = `M ${s.y} ${s.x}
C ${(s.y + d.y) / 2} ${s.x},
${(s.y + d.y) / 2} ${d.x},
${d.y} ${d.x}`
return path
}
function click(d) {
if (d.children) {
d._children = d.children;
d.children = null;
} else {
d.children = d._children;
d._children = null;
}
update(d);
}
}
</script>
</body>
答案 0 :(得分：0)
好的，这样做，
但在检查每个案例时要小心在字符串的两端使用空格。困了我一会儿，
WHERE