如何获取svg元素的坐标?

时间:2013-10-03 08:43:06

标签: javascript dom svg d3.js

我使用d3从相对svg位置绘制一条线,因此想要访问元素本身的坐标。我试过这样的东西(“this”指的是元素)

 .attr("x1", this.x)
 .attr("y1", this.y)

但x1和y1默认为零。获取svg元素坐标的正确方法是什么?

编辑:

这是我的(按要求)我的整个代码。我正在尝试在可折叠树上实现这一点,当我将鼠标悬停在圆形节点上时,我会画一条线。该线的起点应该是我点击的圆圈,终点应该是svg平面上的一个固定点。

<!DOCTYPE html>
<html>
  <head>
    <meta http-equiv="Content-Type" content="text/html;charset=utf-8">
    <script type="text/javascript" src="d3.v3.js"></script>
    <script type="text/javascript" src="d3.layout.js"></script>
    <style type="text/css">

.node circle {
  cursor: pointer;
  fill: #fff;
  stroke: steelblue;
  stroke-width: 1.5px;
}

.node text {
  font: 10px sans-serif;
}

path.link {
  fill: none;
  stroke: #ccc;
  stroke-width: 1.5px;
}

    </style>
  </head>
  <body>
    <div id="chart"></div>
    <script type="text/javascript">

var m = [20, 120, 20, 120],
    w = 2240 - m[1] - m[3],
    h = 800 - m[0] - m[2],
    i = 0,
    duration = 500,
    root;

var tree = d3.layout.tree()
    .size([h, w]);

var diagonal = d3.svg.diagonal()
    .projection(function(d) { return [d.y, d.x]; });

var vis = d3.select("#chart").append("svg")
    .attr("width", w + m[1] + m[3])
    .attr("height", h + m[0] + m[2])
  .append("g")
    .attr("transform", "translate(" + m[3] + "," + m[0] + ")");

function loadData(json) {
    root = json;
    d3.select("#processName").html(root.text);
    root.x0 = h / 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;
    }
 };

// 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);
}

function update(source) {

  // Compute the new tree layout.
  var nodes = tree.nodes(root).reverse();

  // Normalize for fixed-depth.
  nodes.forEach(function(d) { d.y = d.depth * 180; });

  // Update the nodes…
  var node = vis.selectAll("g.node")
      .data(nodes, function(d) { return d.id || (d.id = ++i); });

  // Enter any new nodes at the parent's previous position.
  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("r", 1e-6)
      .style("fill", function(d) { return d._children ? "lightsteelblue" : "#fff"; })
    .on('mouseover', function(d, i){
        var dataset = [1,2,3,4]                                    // HERE
        vis.selectAll("line")                                      // HERE
            .data(dataset)                                 // HERE
            .enter()                                       // HERE
            .append("line")                                // HERE
            .attr("x1", this.x)                            // HERE'S THE PROBLEM FOR PERRY
            .attr("y1", this.y)                            // HERE'S THE PROBLEM FOR PERRY
            .attr("x2", 500)                               // HERE
            .attr("y2", 500)                               // HERE
            .style("stroke", "rgb(6,120,155)");}           // HERE
       ); 

  nodeEnter.append("text")
      .attr("x", function(d) { return d.children || d._children ? -10 : 10; })
      .attr("dy", ".35em")
      .attr("text-anchor", function(d) { return d.children || d._children ? "end" : "start"; })
      .text(function(d) { return d.name; })
      .style("fill-opacity", 1e-6);

  // Transition nodes to their new position.
  var nodeUpdate = node.transition()
      .duration(duration)
      .attr("transform", function(d) { return "translate(" + d.y + "," + d.x + ")"; });

  nodeUpdate.select("circle")
      .attr("r", 4.5)
      .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.y + "," + source.x + ")"; })
      .remove();

  nodeExit.select("circle")
      .attr("r", 1e-6);

  nodeExit.select("text")
      .style("fill-opacity", 1e-6);

  // 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("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;
  });

};

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 ]
});


    </script>
  </body>
</html>

7 个答案:

答案 0 :(得分:54)

element.getBoundingClientRect()方法将返回元素相对于视口的正确坐标,无论svg是否已缩放和/或翻译。

See this question and answer

虽然getBBox()适用于未转换的空间,但如果已将比例和平移应用于布局,则它将不再准确。当平移和缩放生效时,getBoundingClientRect()函数在强制布局项目中对我很有效,我希望将HTML Div元素作为标签附加到节点而不是使用SVG Text元素。

答案 1 :(得分:24)

确定坐标的方法取决于您正在使用的元素。例如,对于circlecxcy属性确定中心位置。此外,您可以通过translation属性应用transform来更改任何坐标的参考点。

一般用于获取屏幕坐标的大多数方法都不适用于SVG。此外,如果要绘制的线与它所连接的元素位于同一容器中,则可能不需要绝对坐标。

编辑:

在您的特定代码中,获取节点的位置非常困难,因为它由父元素的转换确定。因此,您需要获取父节点的transform属性并从中提取转换。

d3.transform(d3.select(this.parentNode).attr("transform")).translate

工作jsfiddle here

答案 2 :(得分:6)

您可以使用函数getBBox()来获取路径的边界框。这将为您提供可包含渲染路径的最紧密矩形的位置和大小。

使用此方法优于读取x和y值的一个优点是它可以与所有图形对象一起使用。有多个对象而不是没有x和y的路径,例如具有cx和cy的圆圈。

答案 3 :(得分:4)

我使用合并功能,如下所示:

 element.transform.baseVal.consolidate()

.e和.f值对应于x和y坐标

答案 4 :(得分:1)

我能像那样处理它;

svg.selectAll("rect")
    .data(zones)
    .enter()
    .append("rect")
    .attr("id", function (d) { return "zone" + d.zone; })
    .attr("class", "zone")
    .attr("x", function (d, i) {
        if (parseInt(i / (wcount)) % 2 == 0) {
            this.xcor = (i % wcount) * zoneW;
        }
        else {
            this.xcor = (zoneW * (wcount - 1)) - ((i % wcount) * zoneW);
        }

        return this.xcor;
    })

,你可以找到x坐标

svg.select("#zone1").on("click",function(){alert(this.xcor});

答案 5 :(得分:0)

我试图选择一个带有矩形的svg区域,并从中获取所有元素。为此,element.getBoundingClientRect()对我来说非常理想。它会返回svg元素的当前坐标,而不管svg是缩放还是转换。

答案 6 :(得分:-1)

svg.selectAll("rect")
.attr('x',function(d,i){
    // get x coord
    console.log(this.getBBox().x, 'or', d3.select(this).attr('x'))
})
.attr('y',function(d,i){
    // get y coord
    console.log(this.getBBox().y)
})
.attr('dx',function(d,i){
    // get dx coord
    console.log(parseInt(d3.select(this).attr('dx')))
})