我正在为繁殖大鼠创建一个谱系树图,我无法计算每个树节点所需的水平距离,因为后代数量变化或是动态的。
我找到了一个用于绘制树的脚本,但看起来它没有任何文档。
var node = new Node({
title: 'M: ' + pair.male + '<br />F: ' + pair.female,
stage: stage,
w: NODE_DIMENSIONS.w,
h: NODE_DIMENSIONS.h,
x: COORDINATES.x + (INCREMENTS.x * pair.column), // Formula should adjust based on descendants - HELP
y: COORDINATES.y + (INCREMENTS.y * pair.generation)
}).attach();
如果您想要使用它并在下面提供完整的代码,那么这里是Fiddle。
<div id="stage"></div>
.node h4 {
position: static;
left: auto;
bottom: auto;
width: 80px;
height: 80px;
text-align: center;
vertical-align: middle;
display: table-cell;
text-shadow: none;
color: #ffffff;
font-weight: bold;
}
var stage = $('#stage');
var NODE_DIMENSIONS = { w: 80, h: 80 };
var SEGMENT_DIMENSIONS = { h: 5 };
var COORDINATES = { x: 50, y: 50 };
var INCREMENTS = { x: 200, y: 150 };
// Sample JSON array resulting from an AJAX call.
var mating = [{
"name": "one",
"male": 1234,
"female": 5643,
"male_lineage": null,
"female_parent": null,
"generation": 0,
"column": 0
}, {
"name": "two",
"male": 6737,
"female": 1627,
"male_lineage": ["four"],
"female_parent": null,
"generation": 0,
"column": 2
}, {
"name": "three",
"male": 9332,
"female": 6227,
"male_lineage": ["five", "six"],
"female_parent": null,
"generation": 0,
"column": 3
}, {
"name": "four",
"male": 1111,
"female": 6537,
"male_lineage": null,
"female_parent": "one",
"generation": 1,
"column": 1
}, {
"name": "five",
"male": 8853,
"female": 3189,
"male_lineage": null,
"female_parent": "two",
"generation": 1,
"column": 2
}, {
"name": "six",
"male": 8853,
"female": 3189,
"male_lineage": null,
"female_parent": "three",
"generation": 1,
"column": 3
}];
var m = new Map();
for (var i = 0; i < mating.length; i++) {
var pair = mating[i];
var node = new Node({
title: 'M: ' + pair.male + '<br />F: ' + pair.female,
stage: stage,
w: NODE_DIMENSIONS.w,
h: NODE_DIMENSIONS.h,
x: COORDINATES.x + (INCREMENTS.x * pair.column), // Formula should adjust based on descendants - HELP
y: COORDINATES.y + (INCREMENTS.y * pair.generation)
}).attach();
var element = {
"pair": pair,
"node": node
};
m.set(pair.name, element);
}
m.forEach(function(element, key, m) {
// We are going to create 2 segments
// First is for the male lineage
if (element.pair.male_lineage != null) {
for(var i = 0; i < element.pair.male_lineage.length; i++) {
new Segment({
h: SEGMENT_DIMENSIONS.h,
stage: stage,
origin: element.node,
destination: m.get(element.pair.male_lineage[i]).node
}).attach();
}
}
// Last is the female parent
if (element.pair.female_parent != null) {
new Segment({
h: SEGMENT_DIMENSIONS.h,
stage: stage,
origin: element.node,
destination: m.get(element.pair.female_parent).node
}).attach();
}
});
答案 0 :(得分:2)
甚至没有尝试(请原谅我......)深入研究“你的”特定问题的原因,我可以明确地说“树木绘图本身就是递归问题。“
此外,不直接映射到HTML“DOM”的问题。(您必须“之后”处理DOM ,之前,你决定“树需要看起来像什么。”)
一般来说,树木绘画问题需要从下到上递归地工作:最低的非叶节点平均分配它们的叶子,然后在它们的孩子的最左边和最右边之间等距离定位,然后报告他们的宽度等于他们孩子宽度的范围。 (这个问题的递归描述然后解决自己,或多或少令人满意,包含整个树。)
一旦你确定了整个树想要如何出现,你的下一个(但是,完全不相关的......)问题是:如何操纵 DOM -tree,“或多或少地驱动浏览器”,实际产生你想要的视觉结果。
(“呃......除了Internet Explorer 8。”)