如何从按钮运行<script>

时间：2017-12-03 02:07:11

标签： javascript d3.js

我是新的网页设计，我正在使用javascript库绘制树。我的问题是它在页面加载时自动绘制。我想有一个按钮，然后点击这幅画，但我找不到怎么做。尝试将该函数传递给javascript文件，但在传递它时我不知道在html文件中要调用它。

＆＃xA;＆＃xA;

这是小提琴 https://jsfiddle.net/nse4w959/

＆＃xA;＆＃xA; 
  ＆lt; script src =“http://d3js.org/d3.v3.min.js”＆gt;＆lt; / script＆gt;＆＃xA;＆lt; script＆gt;＆＃xA; var treeData = [＆＃xA; {＆＃XA; “名称”：“顶级”，＆＃xA; “parent”：“null”，＆＃xA; “孩子们”：[＆＃xA; {＆＃XA; “名称”：“2级：A”，＆＃xA; “父母”：“顶级”，＆＃xA; “孩子们”：[＆＃xA; {＆＃XA; “名字”：“A之子”，＆＃xA; “父母”：“等级2：A”＆＃xA; }，＆＃XA; {＆＃XA; “名字”：“A的女儿”，＆＃xA; “父母”：“等级2：A”＆＃xA; }＆＃XA; ]＆＃XA; }，＆＃XA; {＆＃XA; “名称”：“2级：B”，＆＃xA; “父母”：“顶级”＆＃xA; }＆＃XA; ]＆＃XA; }＆＃XA; ];＆＃XA; // **************生成树形图*****************＆＃xA; var margin = {top：40，right：120，bottom：20，left：120}，＆＃xA; width = 960  -  margin.right  -  margin.left，＆＃xA; height = 500  -  margin.top  -  margin.bottom;＆＃xA; var i = 0;＆＃xA; var tree = d3.layout.tree（）。​​size（[height，width]）;＆＃xA; var diagonal = d3.svg.diagonal（）。projection（function（d）{return [d.x，d.y];}）;＆＃xA; var svg = d3.select（“body”）。append（“svg”）＆＃xA; .attr（“width”，width + margin.right + margin.left）＆＃xA; .attr（“height”，height + margin.top + margin.bottom）.append（“g”）＆＃xA; .attr（“transform”，“translate（”+ margin.left +“，”+ margin.top +“）”）;＆＃xA; root = treeData [0];＆＃xA;更新（根）;＆＃XA;功能更新（来源）{＆＃xA; //计算新的树布局。＆＃xA; var nodes = tree.nodes（root）.reverse（），＆＃xA; links = tree.links（nodes）;＆＃xA; //标准化为fixed-depth.nodes.forEach（function（d）{d.y = d.depth * 100;}）;＆＃xA; //声明节点......＆＃xA; var node = svg.selectAll（“g.node”）。data（nodes，function（d）{return d.id ||（d.id = ++ i）;}）;＆＃xA; //输入节点。＆＃xA; var nodeEnter = node.enter（）。append（“g”）。attr（“class”，“node”）＆＃xA; .attr（“transform”，function（d）{＆＃xA; return“translate（”+ d.x +“，”+ d.y +“）”;}）;＆＃xA; nodeEnter.append（“circle”）。attr（“r”，10）.style（“fill”，“＃ff”）;＆＃xA; nodeEnter.append（“text”）。attr（“y”，function（d）{＆＃xA; return d.children || d._children？-18：18;}）＆＃xA; .attr（“dy”，“。35em”）＆＃xA; .attr（“text-anchor”，“middle”）＆＃xA; .text（function（d）{return d.name;}）＆＃xA; .style（“fill-opacity”，1）;＆＃xA; //声明链接...＆＃xA; var link = svg.selectAll（“path.link”）＆＃xA; .data（links，function（d）{return d.target.id;}）;＆＃xA; //输入链接。＆＃xA; link.enter（）。insert（“path”，“g”）＆＃xA; .attr（“class”，“link”）＆＃xA; .attr（“d”，对角线）;＆＃xA; }＆＃XA;＆LT; /脚本＆GT;＆＃XA;
＆＃XA;

3 个答案:

答案 0 :(得分：2)

调用按钮的最简单方法是

<button onclick="update()">Update</button>

答案 1 :(得分：1)

您可以使用Javascript函数包装代码，并使用按钮的onclick属性将其激活。

检查下面的代码并查看其工作原理

<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="utf-8">
<title>Collapsible Tree Example</title>
<style>
.node circle {
  fill: #fff;
  stroke: steelblue;
  stroke-width: 3px;
}

.node text { font: 12px sans-serif; }

.link {
  fill: none;
  stroke: #ccc;
  stroke-width: 2px;
}
</style>
</head>
<body>

<!-- load the d3.js library --> 
<script src="http://d3js.org/d3.v3.min.js"></script>
<script>

function renderTree() {

var treeData = [
 {
"name": "Top Level",
"parent": "null",
"children": [
  {
    "name": "Level 2: A",
    "parent": "Top Level",
    "children": [
      {
        "name": "Son of A",
        "parent": "Level 2: A"
      },
      {
        "name": "Daughter of A",
        "parent": "Level 2: A"
      }
    ]
  },
  {
    "name": "Level 2: B",
    "parent": "Top Level"
  }
]
}
];
// ************** Generate the tree diagram  *****************
var margin = {top: 40, right: 120, bottom: 20, left: 120},
width = 960 - margin.right - margin.left,
height = 500 - margin.top - margin.bottom;
var i = 0;
var tree = d3.layout.tree().size([height, width]);
var diagonal = d3.svg.diagonal().projection(function(d) { return [d.x, d.y]; });
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 + ")");
root = treeData[0];
update(root);
function update(source) {
 // Compute the new tree layout.
var nodes = tree.nodes(root).reverse(),
links = tree.links(nodes);
// Normalize for fixed-depth.nodes.forEach(function(d) { d.y = d.depth * 100; });
// Declare the nodes…
var node = svg.selectAll("g.node").data(nodes, function(d) { return d.id || (d.id = ++i); });
// Enter the nodes.
var nodeEnter = node.enter().append("g").attr("class", "node")
.attr("transform", function(d) { 
 return "translate(" + d.x + "," + d.y + ")"; });
nodeEnter.append("circle").attr("r", 10).style("fill", "#fff");
nodeEnter.append("text").attr("y", function(d) { 
      return d.children || d._children ? -18 : 18; })
  .attr("dy", ".35em")
  .attr("text-anchor", "middle")
  .text(function(d) { return d.name; })
  .style("fill-opacity", 1);
// Declare the links…
var link = svg.selectAll("path.link")
.data(links, function(d) { return d.target.id; });
 // Enter the links.
 link.enter().insert("path", "g")
  .attr("class", "link")
  .attr("d", diagonal);
}

}
</script>
<input type="button" value="Render" onclick="renderTree();">
</body>
</html>

答案 2 :(得分：0)

https://www.codeproject.com/Tips/1021936/Creating-Vertical-Collapsible-Tree-With-d-js支持

html文件

<html>
<head>
<script src="http://d3js.org/d3.v3.min.js"></script>
</head>
<body>
<input id="clickMe" type="button" value="Click Me" onclick="go();" />
<div id="tree">
</div>
</body>
</html>

css文件

.node circle {
fill: #fff;
stroke: steelblue;
stroke-width: 3px;
}
.node text {
font: 12px sans-serif;
}
.link {
fill: none;
stroke: #ccc;
stroke-width: 2px;
}
.tree {
margin-bottom: 10px;
overflow: auto;
}

js file 

function go(){
var treeData =
  {
      "name": "Top Level",
      "parent": "null",
      "children": [
        {
            "name": "Level 2: A",
            "parent": "Top Level",
            "children": [
              {
                  "name": "Son of A",
                  "parent": "Level 2: A",
                   "children": []
              },
              {
                  "name": "Daughter of A",
                  "parent": "Level 2: A",
                  "children": []
              }
            ]
        },
        {
            "name": "Level 2: B",
            "parent": "Top Level",
            "children": []
        }
      ]
  };
BuildVerticaLTree(treeData,"#tree");
}
function BuildVerticaLTree(treeData, treeContainerDom)      {
  var margin = {
    top: 40,
    right: 120,
    bottom: 20,
    left: 120
  };
  var width = 960 - margin.right - margin.left;
  var height = 960 - margin.top - margin.bottom;

  var i = 0,
    duration = 750;
  var tree = d3.layout.tree()
    .size([height, width]);
  var diagonal = d3.svg.diagonal()
    .projection(function(d) {
      return [d.x, d.y];
    });
  var svg = d3.select(treeContainerDom).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 + ")");
  root = treeData;

  update(root);

  function update(source) {
    // Compute the new tree layout.
    var nodes = tree.nodes(root).reverse(),
      links = tree.links(nodes);
    // Normalize for fixed-depth.
    nodes.forEach(function(d) {
      d.y = d.depth * 100;
    });
    // Declare the nodes…
    var node = svg.selectAll("g.node")
      .data(nodes, function(d) {
        return d.id || (d.id = ++i);
      });
    // Enter the nodes.
    var nodeEnter = node.enter().append("g")
      .attr("class", "node")
      .attr("transform", function(d) {
        return "translate(" + source.x0 + "," + source.y0 + ")";
      }).on("click", nodeclick);
    nodeEnter.append("circle")
      .attr("r", 10)
      .attr("stroke", function(d) {
        return d.children || d._children ? "steelblue" : "#00c13f";
      })
      .style("fill", function(d) {
        return d.children || d._children ? "lightsteelblue" : "#fff";
      });
    //.attr("r", 10)
    //.style("fill", "#fff");
    nodeEnter.append("text")
      .attr("y", function(d) {
        return d.children || d._children ? -18 : 18;
      })
      .attr("dy", ".35em")
      .attr("text-anchor", "middle")
      .text(function(d) {
        return d.name;
      })
      .style("fill-opacity", 1e-6);
    // Transition nodes to their new position.
    //horizontal tree
    var nodeUpdate = node.transition()
      .duration(duration)
      .attr("transform", function(d) {
        return "translate(" + d.x + "," + d.y + ")";
      });
    nodeUpdate.select("circle")
      .attr("r", 10)
      .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.x + "," + source.y + ")";
      })
      .remove();
    nodeExit.select("circle")
      .attr("r", 1e-6);
    nodeExit.select("text")
      .style("fill-opacity", 1e-6);
    // Update the links…
    // Declare the links…
    var link = svg.selectAll("path.link")
      .data(links, function(d) {
        return d.target.id;
      });
    // Enter the links.
    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;
    });
  }

  // Toggle children on click.
  function nodeclick(d) {
    if (d.children) {
      d._children = d.children;
      d.children = null;
    } else {
      d.children = d._children;
      d._children = null;
    }
    update(d);
  }
}
相关问题
最新问题