Javascript D3带矩形和文本换行的水平树布局
多年来一直在努力创建一个带有矩形而不是圆形的水平树布局,并将文本包装在这些矩形中。我所做的一切似乎都不起作用,我已经尝试过了,但无论是谁做的,都遗漏了一个关键步骤,即在行前定义变量Javascript D3带矩形和文本换行的水平树布局,javascript,jquery,html,svg,d3.js,Javascript,Jquery,Html,Svg,D3.js,多年来一直在努力创建一个带有矩形而不是圆形的水平树布局,并将文本包装在这些矩形中。我所做的一切似乎都不起作用,我已经尝试过了,但无论是谁做的,都遗漏了一个关键步骤,即在行前定义变量d if(d.name.length>26)停止整个脚本运行 我还尝试使用d3plus.js from将文本包装在rect标记中,但实际上它一半的时间都不起作用,似乎需要一个类似于单击功能的触发器才能起作用。还考虑将其用作文本包装的指南 在我的研究中,我没有发现有人将水平、矩形和包装文本组合在一个图表中。 另外,我是一
d
if(d.name.length>26)
停止整个脚本运行
我还尝试使用d3plus.js from将文本包装在rect
标记中,但实际上它一半的时间都不起作用,似乎需要一个类似于单击
功能的触发器才能起作用。还考虑将其用作文本包装的指南
在我的研究中,我没有发现有人将水平、矩形和包装文本组合在一个图表中。
另外,我是一个有点D3Noob的,所以所有的帮助是感激的
这是一本书
以下是我正在使用的当前代码,该代码不起作用:
var w = 960,
h = 2000,
i = 0,
duration = 500,
root;
var tree = d3.layout.tree()
.size([h, w - 160]);
var diagonal = d3.svg.diagonal()
.projection(function(d) { return [d.y, d.x]; });
var vis = d3.select("#container").append("svg:svg")
.attr("width", w)
.attr("height", h)
.append("svg:g")
.attr("transform", "translate(40,0)");
root = treeData[0];
root.x0 = h / 2;
root.y0 = 0;
update(root);
function update(source) {
// Compute the new tree layout.
var nodes = tree.nodes(root).reverse();
// Update the nodes…
var node = vis.selectAll("g.node")
.data(nodes, function(d) { return d.id || (d.id = ++i); });
var nodeEnter = node.enter().append("svg:g")
.attr("class", "node")
.attr("transform", function(d) { return "translate(" + source.y0 + "," + source.x0 + ")"; });
// Enter any new nodes at the parent's previous position.
nodeEnter.append("svg:rect")
.attr("width", 150)
.attr("height", function(d) { return (d.name.length > 30) ? 38 : 19;})
.attr("y",-11)
.attr("rx",2)
.attr("ry",2)
.style("fill", function(d) { return d._children ? "lightsteelblue" : "#fff"; })
.on("click", click);
if (d.name.length > 26) {
nodeEnter.append("svg:text")
.attr("x", function(d) { return d._children ? -8 : 8; })
.attr("y", 3)
.text(function(d) { return d.name; });
} else {
nodeEnter.append("svg:text")
.attr("x", function(d) { return d._children ? -8 : 8; })
.attr("y", 3)
.append("svg:tspan")
.text(function(d) { return d.name.slice(0,26); })
.append("svg:tspan")
.attr("x", function(d) { return d._children ? -8 : 8; })
.attr("y",15)
.text(function(d) { return d.name.slice(26); });
}
}
// Transition nodes to their new position.
nodeEnter.transition()
.duration(duration)
.attr("transform", function(d) { return "translate(" + d.y + "," + d.x + ")"; })
.style("opacity", 1)
.select("rect")
.style("fill", "lightsteelblue");
node.transition()
.duration(duration)
.attr("transform", function(d) { return "translate(" + d.y + "," + d.x + ")"; })
.style("opacity", 1);
node.exit().transition()
.duration(duration)
.attr("transform", function(d) { return "translate(" + source.y + "," + source.x + ")"; })
.style("opacity", 1e-6)
.remove();
// Update the links…
var link = vis.selectAll("path.link")
.data(tree.links(nodes), function(d) { return d.target.id; });
// Enter any new links at the parent's previous position.
link.enter().insert("svg:path", "g")
.attr("class", "link")
.attr("d", function(d) {
var o = {x: source.x0, y: source.y0};
return diagonal({source: o, target: o});
})
.transition()
.duration(duration)
.attr("d", diagonal);
// 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;
});
}
// 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);
}
d3.select(self.frameElement).style("height", "2000px");
以及我的json:
var treeData = [
{
"name": "Do trainees require direction as to what to do or how to do the task (either before they start or while they are completing it?",
"children": [
{
"name": "Can they satisfactorily complete the task assigned to them?",
"children": [
{
"name": "Rating level 4",
"parent": "A",
},
{
"name": "How many problems / queries are there that still need to be addressed / resolved to be able to satisfactorily complete the task?",
"children": [
{
"name": "Are problems / queries fundamental to the completion of the task at hand?",
"children": [
{
"name": "Rating level 4",
},
{
"name": "Can the problems be resolved by the trainee (after receiving guidance)?",
"children": [
{
"name": "Rating level 3",
},
{
"name": "Can the problems be resolved by the trainee (after receiving guidance)?",
"children": [
{
"name": "Rating level 2",
},
{
"name": "Rating level 1",
}
]
}
]
}
]
},
{
"name": "Are problems / queries fundamental to the completion of the task at hand?",
}
]
}
]
},
{
"name": "Can they satisfactorily complete the task assigned to them?",
"children": [
{
"name": "Rating 1",
},
{
"name": "Rating 2",
},
{
"name": "Rating 3",
},
{
"name": "Rating 4",
}
]
}
]
}];
您的代码在此行抛出错误:
if (d.name.length > 26) {
未定义d
。当d3
代码引用d时,它通常在数据绑定的范围内。在代码中的这个位置,您没有循环绑定,例如:
nodeEnter.append("text")
.attr("x", function(d) {
return d._children ? -8 : 8;
})
.attr("y", 3)
.attr("dy", "0em")
.text(function(d) {
return d.name; // d is defined from the binding
});
也就是说,我喜欢链接到的wrap
函数。因此,像上面那样添加文本
,然后包装文本:
wrap(d3.selectAll('text'),150);
下面是对包裹的快速修改,可以调整矩形的大小:
function wrap(text, width) {
text.each(function() {
var text = d3.select(this),
words = text.text().split(/\s+/).reverse(),
word,
line = [],
lineNumber = 0,
lineHeight = 1.1, // ems
y = text.attr("y"),
dy = parseFloat(text.attr("dy")),
tspan = text.text(null).append("tspan").attr("x", 0).attr("y", y).attr("dy", dy + "em");
while (word = words.pop()) {
line.push(word);
tspan.text(line.join(" "));
if (tspan.node().getComputedTextLength() > width) {
line.pop();
tspan.text(line.join(" "));
line = [word];
tspan = text.append("tspan").attr("x", 0).attr("y", y).attr("dy", ++lineNumber * lineHeight + dy + "em").text(word);
}
}
// find corresponding rect and reszie
d3.select(this.parentNode.children[0]).attr('height', 19 * (lineNumber+1));
});
}
示例。你可能救了我的命,我很快就要试试。这很有效!有没有一种方法可以缩短初始路径(从原点开始),这样它们就不会延伸那么远,就像这样