我正在使用plotly javascript绘制一些情节。但我真正需要的是网址,而且我并不需要ploly在网上绘制它。
以下是一个示例https://codepen.io/slfan2013/pen/xpWMyW
。我真正需要的是url
,我不希望网站上的情节显示。怎么做?谢谢!
答案 0 :(得分:1)
您不需要实际绘制绘图,可以在Plotly.toImage中传递绘图数据,例如:
var trace1 = {
x: [1, 2, 3, 4],
y: [10, 15, 13, 17],
mode: 'markers'
};
var trace2 = {
x: [2, 3, 4, 5],
y: [16, 5, 11, 10],
mode: 'lines'
};
var trace3 = {
x: [1, 2, 3, 4],
y: [12, 9, 15, 12],
mode: 'lines+markers'
};
var data = [trace1, trace2, trace3];
var layout = {
title: 'Line and Scatter Plot'
};
Plotly.toImage({
data: data,
layout: layout
}, {
height: 600,
width: 800
})
.then(
function(url) {
console.log(url) // this is what i really need!! I dont want it to be ploted~!
}
);
<head>
<!-- Plotly.js -->
<script src="https://cdn.plot.ly/plotly-latest.min.js"></script>
</head>
<body>
<div id="myDiv">
<!-- Plotly chart will be drawn inside this DIV -->
</div>
<script>
<!-- JAVASCRIPT CODE GOES HERE -->
</script>
</body>
让我知道是否有帮助!
答案 1 :(得分:0)
这是我的解决方案,您可以设置div CSS属性C# WebBrowser
,以便不显示图表。然后使用var Jasmine2HtmlReporter = require('protractor-jasmine2-html-reporter');
onPrepare: function () {
jasmine.getEnv().addReporter(
new Jasmine2HtmlReporter({
savePath: __dirname+'/qualityreports/testresults/e2e',
takeScreenshots: true,
takeScreenshotsOnlyOnFailures: true,
filePrefix: 'automationReport',
consolidate: true,
cleanDestination: false,
consolidateAll: true
})
);
},
在获取所需内容后删除生成的图形!
总而言之,图表已创建(但不可见),获取了网址,然后图表被清除!
display:none
plotly.purge()
var trace1 = {
x: [1, 2, 3, 4],
y: [10, 15, 13, 17],
mode: 'markers'
};
var trace2 = {
x: [2, 3, 4, 5],
y: [16, 5, 11, 10],
mode: 'lines'
};
var trace3 = {
x: [1, 2, 3, 4],
y: [12, 9, 15, 12],
mode: 'lines+markers'
};
var data = [trace1, trace2, trace3];
var layout = {
title: 'Line and Scatter Plot'
};
Plotly.newPlot('myDiv', data, layout).then(
function(gd) {
Plotly.toImage(gd, {
height: 600,
width: 800
})
.then(
function(url) {
console.log(url); // this is what i really need!! I dont want it to be ploted~!
Plotly.purge('myDiv');
}
)
});;