Hello,
> function displayMultiple(){
> while (i < 3) {
> x = i*10;
> y = i*10;
> i++;
> display('inline');
> document.getElementById("cross").setAttributeNS
> (null,"transform","translate("+x+" "+y+")");
> setTimeout("display('none')",1000);
> }
> }
This quickly moves ('translate') the same element three times, and
sets three timers. So, at the end you're seeing this element where it
was moved to last and then it gets "display('none')"-ed three times in
a row.
To display the element three times you need to have three instances of
it being rendered. Either use three elements, or consider looking into
the "use" element in SVG
(http://www.w3.org/TR/SVG/struct.html#UseElement). From the SVG spec:
"Any 'svg', 'symbol', 'g', graphics element or other 'use' is
potentially a template object that can be re-used (i.e., "instanced")
in the SVG document via a 'use' element. The 'use' element references
another element and indicates that the graphical contents of that
element is included/drawn at that given point in the document."
--Misha
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