我有一个php变量$blade,我为每个blade创建一个div和图形:
<div id="$blade" style="width:1000px; height:300px;"></div>
g$blade = new Dygraph(document.getElementById("$blade"),"./Data/$node/$blade.txt",{ legend: 'always', title: '$node-$blade', valueRange: [0, 120],visibility: [true, true]});
我将php刀片变量名称保存在javascript变量中:
var blades=[];
foreach ($mabez_blades as &$blade) {
echo <<<AAA
<script type="text/javascript">
blades.push("g$blade");
//console.log(blades);
</script>
AAA;
}
console.log(blades) -> ["gSC_2_1", "gSC_2_2", "gPL_2_3", "gPL_2_4", "gPL_2_5", "gPL_2_6", "gPL_2_7", "gPL_2_8", "gPL_2_9", "gPL_2_10"]
我需要制作一个按钮来将所有这些图形转换为图片并下载它们,问题是我可以直接使用它的名称访问一个图形,我不能动态地迭代图形名称。
<img id="img" style="display:none"/>
<canvas id="canvas" width="1000" height="300" style="display:none" >canvas</canvas>
<script type="text/javascript">
var img = document.getElementById("img");
var canvas = document.getElementById("canvas");
canvas.style.display="none";
var ctx = canvas.getContext("2d");
function saveImage() {
blades.forEach(function(entry) {
var filename = entry+".png";
//Dygraph.Export.asPNG(gSC_2_1, img); //this works for a single graph download
//Dygraph.Export.asPNG(entry, img); //this doesn't work
ctx.clearRect(0, 0, canvas.width, canvas.height);
ctx.drawImage(img, 0, 0);
var link = document.createElement('a');
link.download = filename;
link.href = canvas.toDataURL("image/png").replace("image/png", "image/octet-stream");
link.click();
});
};
</script>
产生以下错误:
Uncaught Type Error: Failed to execute 'drawImage' on 'CanvasRenderingContext2D': The provided value is not of type '(HTMLImageElement or HTMLVideoElement or HTMLCanvasElement or ImageBitmap)'
我找到了这个问题的答案,因为我使用$blade作为div id,它被认为是div元素而不仅仅是字符串。因此,我将图形元素放在一个对象中,键为$blade,值为图形对象,并将div id从$blade更改为div$blade。如果有人有更好的建议,我将非常感谢他/她。
<div id="div$blade" style="width:1000px; height:300px;"></div>
<script type="text/javascript">
var graphs={};
graphs.$blade = new Dygraph(document.getElementById("div$blade"),"./Data/$node/$blade.txt",{ legend: 'always', title: '$node-$blade', valueRange: [0, 120],visibility: [true, true]});
function saveImage() {
for (var key in graphs){
var filename = key+".png";
Dygraph.Export.asPNG(graphs[key], img);
ctx.clearRect(0, 0, canvas.width, canvas.height);
ctx.drawImage(img, 0, 0);
var link = document.createElement('a');
link.download = filename;
link.href = canvas.toDataURL("image/png").replace("image/png", "image/octet-stream");
link.click();
}}