我想把wiggle按钮放在svg下面,而我只是在svg中得到它并覆盖更新按钮,我无法弄清楚,帮助我,pleaz.Sorry为我的网络提前确认:( ,代码来了:
<!DOCTYPE html>
<meta charset="utf-8">
<title>Streamgraph</title>
<style>
body {
font-family: "Helvetica Neue", Helvetica, Arial, sans-serif;
margin: auto;
position: relative;
width: 960px;
}
button {
position: absolute;
right: 10px;
top: 10px;
}
</style>
<button onclick="transition()">Update</button>
<script src="http://d3js.org/d3.v3.min.js"></script>
<script>
var n = 20, // number of layers
m = 200, // number of samples per layer
zero_stack = d3.layout.stack().offset("zero"),
wig_stack = d3.layout.stack().offset("wiggle"),
exp_stack = d3.layout.stack().offset("expand"),
sil_stack = d3.layout.stack().offset("silhouette"),
layers0 = zero_stack(d3.range(n).map(function() { return bumpLayer(m); })),
layers1 = zero_stack(d3.range(n).map(function() { return bumpLayer(m); }));
var width = 960,
height = 500;
var x = d3.scale.linear()
.domain([0, m - 1])
.range([0, width]);
var y = d3.scale.linear()
.domain([0, d3.max(layers0.concat(layers1), function(layer) { return d3.max(layer, function(d) { return d.y0 + d.y; }); })])
.range([height, 0]);
var color = d3.scale.linear()
.range(["#aad", "#556"]);
var area = d3.svg.area()
.x(function(d) { return x(d.x); })
.y0(function(d) { return y(d.y0); })
.y1(function(d) { return y(d.y0 + d.y); });
var svg = d3.select("body").append("svg")
.attr("width", width)
.attr("wmode","Transparent")
.attr("height", height);
addElem(svg,"rect").attr({
"width" : width,
"pointer-events" : "all",
"float" : "left",
"height" : height
})
.style("visibility","hidden");
svg.selectAll("path")
.data(layers0)
.enter().append("path")
.attr("d", area)
.style("fill", function() { return color(Math.random()); });
d3.select("body")append("button").text("wiggle").attr("float","left").on("click",changewiggle);
function transition() {
d3.selectAll("path")
.data(function() {
var d = layers1;
layers1 = layers0;
return layers0 = d;
})
.transition()
.duration(2500)
.attr("d", area);
}
// Inspired by Lee Byron's test data generator.
function bumpLayer(n) {
function bump(a) {
var x = 1 / (.1 + Math.random()),
y = 2 * Math.random() - .5,
z = 10 / (.1 + Math.random());
for (var i = 0; i < n; i++) {
var w = (i / n - y) * z;
a[i] += x * Math.exp(-w * w);
}
}
var a = [], i;
for (i = 0; i < n; ++i) a[i] = 0;
for (i = 0; i < 5; ++i) bump(a);
return a.map(function(d, i) { return {x: i, y: Math.max(0, d)}; });
}
function changewiggle(){
log("wig");
layers0 = wig_stack(d3.range(n).map(function() { return bumpLayer(m); })),
layers1 = wig_stack(d3.range(n).map(function() { return bumpLayer(m); }));
}
</script>
如果可以的话,请告诉我如何在不点击更新按钮的情况下应用布局来摆动,现在的情况是我点击了摆动按钮,然后我必须点击更新按钮才能看到图表更新其图表。
答案 0 :(得分:2)
我对你想要的一切并不完全清楚,但经过一些调整后,这里是FIDDLE。首先,我添加了图表div以帮助定位摆动按钮。
d3.select("#chart")
.append("button")
.text("wiggle")
.attr("float", "left")
.on("click", changewiggle);
但还有其他一些调整,例如删除导致错误的addElem()部分,从wiggle按钮代码调用转换等。无论如何,让我们看看这是否有帮助。