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Table of Contents
1. Create a tree structure
2. Simulate the radius calculation of each layer of the 3D tree structure under 2D topology
3. 加入z軸坐標(biāo),呈現(xiàn)3D下的樹(shù)狀結(jié)構(gòu)
Home Web Front-end H5 Tutorial Detailed explanation of graphic and text code for 3D network topology tree presentation based on HTML5

Detailed explanation of graphic and text code for 3D network topology tree presentation based on HTML5

Mar 07, 2017 pm 03:36 PM

In HT for Web, both 2D and 3D applications support the display of tree structure data, with different display effects. The tree structure on 2D has an obvious hierarchical relationship, but if the amount of data is large, , it seems not so intuitive, and it is more difficult to find the specified node. However, the tree structure in 3D will appear more intuitive when combined with the elastic layout component of HT for Web. You can see the entire tree structure data at a glance. A rough idea, but under the influence of elastic layout, the hierarchical structure is not so clear. So at this time, the need for a 3D tree with a clear structure comes. So what exactly does this 3D tree look like? Let’s see it together~

To achieve this Effect, where to start? Next, we will break this problem down into several small problems to solve.

1. Create a tree structure

Those who have learned about HT for Web should be familiar with the creation of tree structure data, so I will not discuss it in depth here. . The creation of tree structure data is very simple. In order to make the code more concise, I have encapsulated three methods to create tree structure data. The specific code is as follows:

/**
 * 創(chuàng)建連線
 * @param {ht.DataModel} dataModel - 數(shù)據(jù)容器
 * @param {ht.Node} source - 起點(diǎn)
 * @param {ht.Node} target - 終點(diǎn)
 */
function createEdge(dataModel, source, target) {
    // 創(chuàng)建連線,鏈接父親節(jié)點(diǎn)及孩子節(jié)點(diǎn)
    var edge = new ht.Edge();
    edge.setSource(source);
    edge.setTarget(target);
    dataModel.add(edge);
}

/**
 * 創(chuàng)建節(jié)點(diǎn)對(duì)象
 * @param {ht.DataModel} dataModel - 數(shù)據(jù)容器
 * @param {ht.Node} [parent] - 父親節(jié)點(diǎn)
 * @returns {ht.Node} 節(jié)點(diǎn)對(duì)象
 */
function createNode(dataModel, parent) {
    var node = new ht.Node();
    if (parent) {
        // 設(shè)置父親節(jié)點(diǎn)
        node.setParent(parent);

        createEdge(dataModel, parent, node);
    }
    // 添加到數(shù)據(jù)容器中
    dataModel.add(node);
    return node;
}

/**
 * 創(chuàng)建結(jié)構(gòu)樹(shù)
 * @param {ht.DataModel} dataModel - 數(shù)據(jù)容器
 * @param {ht.Node} parent - 父親節(jié)點(diǎn)
 * @param {Number} level - 深度
 * @param {Array} count - 每層節(jié)點(diǎn)個(gè)數(shù)
 * @param {function(ht.Node, Number, Number)} callback - 回調(diào)函數(shù)(節(jié)點(diǎn)對(duì)象,節(jié)點(diǎn)對(duì)應(yīng)的層級(jí),節(jié)點(diǎn)在層級(jí)中的編號(hào))
 */
function createTreeNodes(dataModel, parent, level, count, callback) {
    level--;
    var num = (typeof count === 'number' ? count : count[level]);

    while (num--) {
        var node = createNode(dataModel, parent);
        // 調(diào)用回調(diào)函數(shù),用戶可以在回調(diào)里面設(shè)置節(jié)點(diǎn)相關(guān)屬性
        callback(node, level, num);
        if (level === 0) continue;
        // 遞歸調(diào)用創(chuàng)建孩子節(jié)點(diǎn)
        createTreeNodes(dataModel, node, level, count, callback);
    }
}

Hehe, the code may be a bit complicated to write. , the simple way is to nest several for loops to create tree-structured data. I won’t go into details here. Next, let’s explore the second question.

2. Simulate the radius calculation of each layer of the 3D tree structure under 2D topology

The biggest problem with the tree structure under 3D is that the level of each node and the level of each layer A node's radius is calculated around its parent node. Now that the tree structure data is available, it is time to start calculating the radius. We start from the two-layer tree structure:

I now create two In the layered tree structure, all child nodes are lined up and do not surround their parent nodes. So how do we determine the positions of these child nodes?

First of all, we need to know that each end node has a circle of its own domain, otherwise there will be overlap between nodes, so here, we assume that the domain radius of the end node is 25, Then the shortest distance between two adjacent nodes will be twice the node field radius, which is 50, and these end nodes will evenly surround their parent nodes, then the opening angle of the two adjacent nodes can be confirmed Come out, with the opening angle and the distance between two points, the shortest radius of a node around its parent node can also be calculated. Suppose the opening angle is a and the minimum distance between two points is b, then the minimum radius r The calculation formula is:

r = b / 2 / sin(a / 2);

Then let’s lay out the tree. The code is written like this:

/**
 * 布局樹(shù)
 * @param {ht.Node} root - 根節(jié)點(diǎn)
 * @param {Number} [minR] - 末端節(jié)點(diǎn)的最小半徑
 */
function layout(root, minR) {
    // 設(shè)置默認(rèn)半徑
    minR = (minR == null ? 25 : minR);
    // 獲取到所有的孩子節(jié)點(diǎn)對(duì)象數(shù)組
    var children = root.getChildren().toArray();
    // 獲取孩子節(jié)點(diǎn)個(gè)數(shù)
    var len = children.length;
    // 計(jì)算張角
    var degree = Math.PI * 2 / len;
    // 根據(jù)三角函數(shù)計(jì)算繞父親節(jié)點(diǎn)的半徑
    var sin = Math.sin(degree / 2),
        r = minR / sin;
    // 獲取父親節(jié)點(diǎn)的位置坐標(biāo)
    var rootPosition = root.p();

    children.forEach(function(child, index) {
        // 根據(jù)三角函數(shù)計(jì)算每個(gè)節(jié)點(diǎn)相對(duì)于父親節(jié)點(diǎn)的偏移量
        var s = Math.sin(degree * index),
            c = Math.cos(degree * index),
            x = s * r,
            y = c * r;

        // 設(shè)置孩子節(jié)點(diǎn)的位置坐標(biāo)
        child.p(x + rootPosition.x, y + rootPosition.y);
    });
}

In the code, you will find that I set the end radius to 25 by default. In this way, we can lay out the structure tree by calling the layout() method. The layout effect is as follows:

It can be seen from the renderings that the default radius of the end node is not very ideal. The layout effect is almost invisible, so we can increase the default radius of the end node to solve the problem. The problem of too dense layout, for example, the effect of setting the default radius to 40 is as follows:

Now that the two-layer tree distribution is solved, let’s take a look at the three-layer How to deal with the tree distribution.

Considering the second layer and the third layer as a whole, the tree structure of the three layers is actually the same as the two layers. The difference is that when processing the second layer nodes, they should be regarded as To process a two-layer tree structure, it is best to use recursion to process this kind of regularity, so we will slightly modify the code and see how it works:

No, the nodes are all overlapping. It seems that simple recursion will not work. So where is the specific problem?

After careful analysis, I found that the domain radius of the father node is determined by the domain radius of its child node. Therefore, you need to know the domain radius of your own node during layout, and the position of the node depends on the domain of the father node. Radius and position information, so that the node position cannot be laid out while calculating the radius.

Now we can only separate the calculation of the radius from the layout and do a two-step operation. Let’s first analyze the calculation of the node radius:

First we need to clarify the most critical conditions, father The radius of a node depends on the radius of its child node. This condition tells us that the node radius can only be calculated from bottom to top. Therefore, the recursive function we design must be recursive first and then calculated. Without further ado, let’s take a look at the specifics. Code:

/**
 * 就按節(jié)點(diǎn)領(lǐng)域半徑
 * @param {ht.Node} root - 根節(jié)點(diǎn)對(duì)象
 * @param {Number} minR - 最小半徑
 */
function countRadius(root, minR) {
    minR = (minR == null ? 25 : minR);

    // 若果是末端節(jié)點(diǎn),則設(shè)置其半徑為最小半徑
    if (!root.hasChildren()) {
        root.a('radius', minR);
        return;
    }

    // 遍歷孩子節(jié)點(diǎn)遞歸計(jì)算半徑
    var children = root.getChildren();
    children.each(function(child) {
        countRadius(child, minR);
    });

    var child0 = root.getChildAt(0);
    // 獲取孩子節(jié)點(diǎn)半徑
    var radius = child0.a('radius');

    // 計(jì)算子節(jié)點(diǎn)的1/2張角
    var degree = Math.PI / children.size();
    // 計(jì)算父親節(jié)點(diǎn)的半徑
    var pRadius = radius / Math.sin(degree);

    // 設(shè)置父親節(jié)點(diǎn)的半徑及其孩子節(jié)點(diǎn)的布局張角
    root.a('radius', pRadius);
    root.a('degree', degree * 2);
}

OK,半徑的計(jì)算解決了,那么接下來(lái)就該解決布局問(wèn)題了,布局樹(shù)狀結(jié)構(gòu)數(shù)據(jù)需要明確:孩子節(jié)點(diǎn)的坐標(biāo)位置取決于其父親節(jié)點(diǎn)的坐標(biāo)位置,因此布局的遞歸方式和計(jì)算半徑的遞歸方式不同,我們需要先布局父親節(jié)點(diǎn)再遞歸布局孩子節(jié)點(diǎn),具體看看代碼吧:

/**
 * 布局樹(shù)
 * @param {ht.Node} root - 根節(jié)點(diǎn)
 */
function layout(root) {
    // 獲取到所有的孩子節(jié)點(diǎn)對(duì)象數(shù)組
    var children = root.getChildren().toArray();
    // 獲取孩子節(jié)點(diǎn)個(gè)數(shù)
    var len = children.length;
    // 計(jì)算張角
    var degree = root.a('degree');
    // 根據(jù)三角函數(shù)計(jì)算繞父親節(jié)點(diǎn)的半徑
    var r = root.a('radius');
    // 獲取父親節(jié)點(diǎn)的位置坐標(biāo)
    var rootPosition = root.p();

    children.forEach(function(child, index) {
        // 根據(jù)三角函數(shù)計(jì)算每個(gè)節(jié)點(diǎn)相對(duì)于父親節(jié)點(diǎn)的偏移量
        var s = Math.sin(degree * index),
            c = Math.cos(degree * index),
            x = s * r,
            y = c * r;

        // 設(shè)置孩子節(jié)點(diǎn)的位置坐標(biāo)
        child.p(x + rootPosition.x, y + rootPosition.y);

        // 遞歸調(diào)用布局孩子節(jié)點(diǎn)
        layout(child);
    });
}

代碼寫完了,接下來(lái)就是見(jiàn)證奇跡的時(shí)刻了,我們來(lái)看看效果圖吧:

不對(duì)呀,代碼應(yīng)該是沒(méi)問(wèn)題的呀,為什么顯示出來(lái)的效果還是會(huì)重疊呢?不過(guò)仔細(xì)觀察我們可以發(fā)現(xiàn)相比上個(gè)版本的布局會(huì)好很多,至少這次只是末端節(jié)點(diǎn)重疊了,那么問(wèn)題出在哪里呢?

不知道大家有沒(méi)有發(fā)現(xiàn),排除節(jié)點(diǎn)自身的大小,倒數(shù)第二層節(jié)點(diǎn)與節(jié)點(diǎn)之間的領(lǐng)域是相切的,那么也就是說(shuō)節(jié)點(diǎn)的半徑不僅和其孩子節(jié)點(diǎn)的半徑有關(guān),還與其孫子節(jié)點(diǎn)的半徑有關(guān),那我們把計(jì)算節(jié)點(diǎn)半徑的方法改造下,將孫子節(jié)點(diǎn)的半徑也考慮進(jìn)去再看看效果如何,改造后的代碼如下:

/**
 * 就按節(jié)點(diǎn)領(lǐng)域半徑
 * @param {ht.Node} root - 根節(jié)點(diǎn)對(duì)象
 * @param {Number} minR - 最小半徑
 */
function countRadius(root, minR) {
   ……

    var child0 = root.getChildAt(0);
    // 獲取孩子節(jié)點(diǎn)半徑
    var radius = child0.a('radius');

    var child00 = child0.getChildAt(0);
    // 半徑加上孫子節(jié)點(diǎn)半徑,避免節(jié)點(diǎn)重疊
    if (child00) radius += child00.a('radius');

   ……
}

下面就來(lái)看看效果吧~

哈哈,看來(lái)我們分析對(duì)了,果然就不再重疊了,那我們來(lái)看看再多一層節(jié)點(diǎn)會(huì)是怎么樣的壯觀場(chǎng)景呢?

哦,NO!這不是我想看到的效果,又重疊了,好討厭。

不要著急,我們?cè)賮?lái)仔細(xì)分析分析下,在前面,我們提到過(guò)一個(gè)名詞——領(lǐng)域半徑,什么是領(lǐng)域半徑呢?很簡(jiǎn)單,就是可以容納下自身及其所有孩子節(jié)點(diǎn)的最小半徑,那么問(wèn)題就來(lái)了,末端節(jié)點(diǎn)的領(lǐng)域半徑為我們指定的最小半徑,那么倒數(shù)第二層的領(lǐng)域半徑是多少呢?并不是我們前面計(jì)算出來(lái)的半徑,而應(yīng)該加上末端節(jié)點(diǎn)自身的領(lǐng)域半徑,因?yàn)樗鼈冎g存在著包含關(guān)系,子節(jié)點(diǎn)的領(lǐng)域必須包含于其父親節(jié)點(diǎn)的領(lǐng)域中,那我們?cè)诳纯瓷蠄D,是不是感覺(jué)末端節(jié)點(diǎn)的領(lǐng)域被侵占了。那么我們前面計(jì)算出來(lái)的半徑代表著什么呢?前面計(jì)算出來(lái)的半徑其實(shí)代表著孩子節(jié)點(diǎn)的布局半徑,在布局的時(shí)候是通過(guò)該半徑來(lái)布局的。

OK,那我們來(lái)總結(jié)下,節(jié)點(diǎn)的領(lǐng)域半徑是其下每層節(jié)點(diǎn)的布局半徑之和,而布局半徑需要根據(jù)其孩子節(jié)點(diǎn)個(gè)數(shù)及其領(lǐng)域半徑共同決定。

好了,我們現(xiàn)在知道問(wèn)題的所在了,那么我們的代碼該如何去實(shí)現(xiàn)呢?接著往下看:

/**
 * 就按節(jié)點(diǎn)領(lǐng)域半徑及布局半徑
 * @param {ht.Node} root - 根節(jié)點(diǎn)對(duì)象
 * @param {Number} minR - 最小半徑
 */
function countRadius(root, minR) {
    minR = (minR == null ? 25 : minR);

    // 若果是末端節(jié)點(diǎn),則設(shè)置其布局半徑及領(lǐng)域半徑為最小半徑
    if (!root.hasChildren()) {
        root.a('radius', minR);
        root.a('totalRadius', minR);
        return;
    }

    // 遍歷孩子節(jié)點(diǎn)遞歸計(jì)算半徑
    var children = root.getChildren();
    children.each(function(child) {
        countRadius(child, minR);
    });

    var child0 = root.getChildAt(0);
    // 獲取孩子節(jié)點(diǎn)半徑
    var radius = child0.a('radius'),
        totalRadius = child0.a('totalRadius');

    // 計(jì)算子節(jié)點(diǎn)的1/2張角
    var degree = Math.PI / children.size();
    // 計(jì)算父親節(jié)點(diǎn)的布局半徑
    var pRadius = totalRadius / Math.sin(degree);

    // 緩存父親節(jié)點(diǎn)的布局半徑
    root.a('radius', pRadius);
    // 緩存父親節(jié)點(diǎn)的領(lǐng)域半徑
    root.a('totalRadius', pRadius + totalRadius);
    // 緩存其孩子節(jié)點(diǎn)的布局張角
    root.a('degree', degree * 2);
}

在代碼中我們將節(jié)點(diǎn)的領(lǐng)域半徑緩存起來(lái),從下往上一層一層地疊加上去。接下來(lái)我們一起驗(yàn)證其正確性:

搞定,就是這樣子了,2D拓?fù)渖厦娴牟季指愣?,那么接下?lái)該出動(dòng)3D拓?fù)淅瞺

3. 加入z軸坐標(biāo),呈現(xiàn)3D下的樹(shù)狀結(jié)構(gòu)

3D拓?fù)渖厦娌季譄o(wú)非就是多加了一個(gè)坐標(biāo)系,而且這個(gè)坐標(biāo)系只是控制節(jié)點(diǎn)的高度而已,并不會(huì)影響到節(jié)點(diǎn)之間的重疊,所以接下來(lái)我們來(lái)改造下我們的程序,讓其能夠在3D上正常布局。

也不需要太大的改造,我們只需要修改下布局器并且將2D拓?fù)浣M件改成3D拓?fù)浣M件就可以了。

/**
 * 布局樹(shù)
 * @param {ht.Node} root - 根節(jié)點(diǎn)
 */
function layout(root) {
    // 獲取到所有的孩子節(jié)點(diǎn)對(duì)象數(shù)組
    var children = root.getChildren().toArray();
    // 獲取孩子節(jié)點(diǎn)個(gè)數(shù)
    var len = children.length;
    // 計(jì)算張角
    var degree = root.a('degree');
    // 根據(jù)三角函數(shù)計(jì)算繞父親節(jié)點(diǎn)的半徑
    var r = root.a('radius');
    // 獲取父親節(jié)點(diǎn)的位置坐標(biāo)
    var rootPosition = root.p3();

    children.forEach(function(child, index) {
        // 根據(jù)三角函數(shù)計(jì)算每個(gè)節(jié)點(diǎn)相對(duì)于父親節(jié)點(diǎn)的偏移量
        var s = Math.sin(degree * index),
            c = Math.cos(degree * index),
            x = s * r,
            z = c * r;

        // 設(shè)置孩子節(jié)點(diǎn)的位置坐標(biāo)
        child.p3(x + rootPosition[0], rootPosition[1] - 100, z + rootPosition[2]);

        // 遞歸調(diào)用布局孩子節(jié)點(diǎn)
        layout(child);
    });
}

上面是改造成3D布局后的布局器代碼,你會(huì)發(fā)現(xiàn)和2D的布局器代碼就差一個(gè)坐標(biāo)系的的計(jì)算,其他的都一樣,看下在3D上布局的效果:

恩,有模有樣的了,在文章的開(kāi)頭,我們可以看到每一層的節(jié)點(diǎn)都有不同的顏色及大小,這些都是比較簡(jiǎn)單,在這里我就不做深入的講解,具體的代碼實(shí)現(xiàn)如下:

var level = 4,
    size = (level + 1) * 20;

var root = createNode(dataModel);
root.setName('root');
root.p(100, 100);

root.s('shape3d', 'sphere');
root.s('shape3d.color', randomColor());
root.s3(size, size, size);

var colors = {},
    sizes = {};
createTreeNodes(dataModel, root, level - 1, 5, function(data, level, num) {
    if (!colors[level]) {
        colors[level] = randomColor();
        sizes[level] = (level + 1) * 20;
    }

    size = sizes[level];

    data.setName('item-' + level + '-' + num);
    // 設(shè)置節(jié)點(diǎn)形狀為球形
    data.s('shape3d', 'sphere');
    data.s('shape3d.color', colors[level]);
    data.s3(size, size, size);
});

在這里引入了一個(gè)隨機(jī)生成顏色值的方法,對(duì)每一層隨機(jī)生成一種顏色,并將節(jié)點(diǎn)的形狀改成了球形,讓頁(yè)面看起來(lái)美觀些(其實(shí)很丑)。

提個(gè)外話,節(jié)點(diǎn)上可以貼上圖片,還可以設(shè)置文字的朝向,可以根據(jù)用戶的視角動(dòng)態(tài)調(diào)整位置,等等一系列的拓展,這些大家都可以去嘗試,相信都可以做出一個(gè)很漂亮的3D樹(shù)出來(lái)。

到此,整個(gè)Demo的制作就結(jié)束了,今天的篇幅有些長(zhǎng),感謝大家的耐心閱讀,在設(shè)計(jì)上或則是表達(dá)上有什么建議或意見(jiàn)歡迎大家提出,點(diǎn)擊這里可以訪問(wèn)HT for Web官網(wǎng)上的手冊(cè)。

The above is the detailed explanation of the graphic code of the 3D network topology tree based on HTML5. For more related content, please pay attention to the PHP Chinese website (www.miracleart.cn)!



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