使用safari在画布中绘制包含html的svg

Question

我正在尝试为网站创建一些动态创建页面的缩略图,我已经通过在svg中添加html找到了解决方案,然后我在画布内的图像上绘制,我在绘制图像后调整大小.

这个解决方案适用于firefox和chrome,但不适用于safari,svg似乎没有被绘制我只是得到一个空白页面.即使我把一些尝试捕获,我也没有得到任何错误,我无法在网上找到解决方案.

我的html只是一个带有画布的基本测试页面,我试图在头上添加一个元标记

<meta http-equiv="content-type" content="application/xhtml+xml; charset=utf-8" />

但它也没有用.

这是我写的代码:

var canvas = document.getElementById('canvasTest'),
            context = canvas.getContext('2d');

            var data = "<svg xmlns='http://www.w3.org/2000/svg' width='200' height='200'>" +
                 "<foreignObject width='100%' height='100%'>" +
                   "<div xmlns='http://www.w3.org/1999/xhtml' style='font-size:40px'>" +
                     "<div>TEST<span>TEST</span></div>" +
                   "</div>" +
                 "</foreignObject>" +
               "</svg>"
            ;

            var DOMURL = self.URL || self.webkitURL || self;
            var img = new Image();                                                                              

            try {
                var svg = new Blob([data], {type: "image/svg+xml;charset=utf-8"});                      
            } catch(e) {                    
                window.BlobBuilder = window.BlobBuilder || window.WebKitBlobBuilder || window.MozBlobBuilder || window.MSBlobBuilder;
                console.log(window.BlobBuilder);

                if(e.name == 'TypeError' && window.BlobBuilder){
                    var bb = new BlobBuilder();
                    bb.append(data);
                    var svg = bb.getBlob("image/svg+xml;charset=utf-8");
                } else if(e.name == "InvalidStateError") {                      
                    var svg = new Blob(data, {type : "image/svg+xml;charset=utf-8"});
                } else {

                }
            }               

            var url = DOMURL.createObjectURL(svg);              

            img.onload = function() {                   
                    context.drawImage(img, 100, 100);                                       
            };

            img.src = url;
            DOMURL.revokeObjectURL(url);

            context.translate(canvas.width / 2, canvas.height / 2);
            context.scale(0.4, 0.4);

Safari可以使用Blob构造函数,因此它不会进入catch.似乎问题是当我传递svg引用(img.src = url)但我无法弄清楚要做什么不同的事情,如果有人有一些想法或解决方案它会很棒.

编辑:safari的控制台输出

Blob:Blob {size = 232,type ="image/svg + xml; charset = utf-8",webkitSlice}

网址:blob:http://myPage.com/96fb502f-46bb-492b-af35-5313bb​​39bd31

Answer 1

基本上解决方案是将mime类型(data:image/svg + xml)放在包含svg的变量中,这样你就不需要再使用blob了,然后在绘制之前用svg设置img src到画布.

var canvas = document.getElementById('canvasTest'),
            context = canvas.getContext('2d');

            var data = "data:image/svg+xml,"+"<svg xmlns='http://www.w3.org/2000/svg' width='200' height='200'>" +
                 "<foreignObject width='100%' height='100%'>" +
                   "<div xmlns='http://www.w3.org/1999/xhtml' style='font-size:40px'>" +
                     "<div xmlns='http://www.w3.org/1999/xhtml'>aaaa<span>aaaa</span></div>" +
                   "</div>" +
                 "</foreignObject>" +
               "</svg>"
            ;

            var img = new Image();

            img.src = data;

            img.onload = function() {
                context.drawImage(img, 100, 100);                   
            };

            context.translate(canvas.width / 2, canvas.height / 2);
            context.scale(0.4, 0.4);