Hi Matthew,

Couldn't you just encapsulate the inner *<svg>* elements in an outer *<div>* 
set to 100% size, and have your inner *<svg>* work relative to that?

Hegart.


On Friday, 11 March 2016 08:55:30 UTC+13, Matthew Lauber wrote:
>
> I'm actually using those attributes on the outer element to position the 
> inner element x% and y% releative to the outer svg.  However, I will 
> experiment if there's some way I can otherwise position them.
>
> On Thursday, March 10, 2016 at 2:27:37 PM UTC-5, Eric Shulman wrote:
>>
>>
>> On Thursday, March 10, 2016 at 9:34:22 AM UTC-8, Matthew Lauber wrote:
>>>
>>> Anyone have a easy solution to using the $link element inside a <svg>? 
>>>  Attached I've got my testing example.  An empty <a> tag works just fine, 
>>> but using <$link> in its place causes the bow not to render.  I't possible 
>>> I've missed something obvious, but I've been banging my head against the 
>>> wall.
>>>
>>
>> Your svg definitions have two *nested* <svg>...</svg> elements.  You have:
>>
>> <svg ...>
>>    <a>
>>       <svg ...>
>>       ...
>>       </svg>
>>    </a>
>> </svg>
>>
>> and
>>
>> <svg ...>
>>    <$link ...>
>>       <svg ...>
>>       ...
>>       </svg>
>>    </$link>
>> </svg>
>>
>> If you remove the outer <svg></svg> pair, the $link widget and the <a> 
>> element work as expected (i.e., creates a link to a tiddler or an href, 
>> respectively).  It seems the the only thing the outer <svg> element adds is 
>> width="100%" height="100%", but you should be able to simply move those 
>> attributes to the inner <svg> so the image will appear the same size as 
>> before.
>>
>> enjoy,
>> -e
>> Eric Shulman
>> TiddlyTools / ELS Design Studios
>> InsideTiddlyWiki: The Missing Manuals
>>
>

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