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Improve style handling in SvgExporter.

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Jürg Lehni 2012-11-05 09:05:32 -08:00
parent 61531949b0
commit 65f1e5c66d
2 changed files with 9 additions and 6 deletions
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2
examples/SVG Export/Rect and Attribute Testing.html
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@ -22,7 +22,7 @@
path1.id = "square1";
path1.strokeCap = "square";
path1.opacity = ".1";
path1.dashArray = "[5, 2]";
path1.dashArray = [5, 2];
path1.dashOffset = "0";
var point2 = new Point(75, 75);

13
src/svg/SvgExporter.js
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@ -39,6 +39,7 @@ var SvgExporter = this.SvgExporter = new function() {
children = group._children;
for (var i = 0, l = children.length; i < l; i++)
svg.appendChild(SvgExporter.exportItem(children[i]));
applyStyle(group, svg);
return svg;
}
@ -339,7 +340,8 @@ var SvgExporter = this.SvgExporter = new function() {
function applyStyle(item, svg) {
var attrs = {},
style = item._style,
parentStyle = item.getParent()._style,
parent = item.getParent(),
parentStyle = parent && parent._style,
properties = {
fillColor: 'fill',
strokeColor: 'stroke',
@ -361,14 +363,15 @@ var SvgExporter = this.SvgExporter = new function() {
// Get a given style only if it differs from the value on the parent
// (A layer or group which can have style values in SVG).
var getter = 'get' + Base.capitalize(name),
value = style[getter](),
parent = parentStyle[getter]();
if (!Base.equals(parent, value)) {
attrs[svgName] = value && /Color$/.test(name)
value = style[getter]();
if (value != null && (!parentStyle
|| !Base.equals(parentStyle[getter](), value))) {
value = /Color$/.test(name)
? value.toCssString()
: name == 'dashArray'
? value.join(',')
: value;
attrs[svgName] = value;
}
});