Drew,

You are right that there would be a problem referencing the parents if they
are not globally unique.  A reference to a parent would have to use a fully
qualified path starting from the root instead of just using the non-unique
parent id.  This could be done, but is the added complexity worth it?

dan

On Wed, Jan 2, 2013 at 5:40 PM, asgallant <[email protected]> wrote:

> All nodes with children must have globally unique id's, or else their
> child nodes won't be able to distinguish which node is their parent.  Since
> trees are programmatically extensible, it makes sense to enforce unique
> id's on all nodes, regardless of whether or not they have any child nodes
> at draw time (as a previously child-less node may have children on redraw).
>  The solution proposed by tay is probably the best way to handle the
> problem, absent significant modifications to the TreeMap code that
> determines the parent-child relationships.
>
>
> On Wednesday, January 2, 2013 4:36:03 PM UTC-5, Daniel LaLiberte wrote:
>
>> Ribardiere ,
>>
>> I don't know of a reason that the ids have to be globally unique, other
>> than it was assumed to be a reasonable assumption, so I agree with your
>> assessment.  It doesn't seem so difficult to fix, but I won't know until I
>> have time to look into it.  We might have to construct global unique ids
>> for every node by using all the ancestors, for example.  Thanks for your
>> posting about this issue.
>>
>> dan
>>
>> On Wed, Jan 2, 2013 at 3:45 PM, Ribardiere Olivier <
>> [email protected]> wrote:
>>
>>> Hi,
>>>
>>> In a Treemap chart, 2 nodes should be able to have the same name if they
>>> don't have the same parent.
>>>
>>> function drawVisualization() {
>>>   // Create and populate the data table.
>>>   var data = google.visualizat**ion.arrayToDataTable([
>>>     ['Child' , 'Parent', 'Size',** 'Color'],
>>>     ['Global',    null,                0,
>>> 0],
>>>     ['Child1',   'Global',             0,
>>> 0],
>>>     ['Child2',   'Global',             0,
>>> 0],
>>>     ['ChildA',   'Child1',            10,
>>> 0],
>>>     ['ChildB',   'Child1',            10,
>>> 0],
>>>     ['ChildA',   'Child2',            10,
>>> 0],
>>>     ['ChildB',   'Child2',            10,
>>> 0]
>>>   ]);
>>>
>>>   // Create and draw the visualization.
>>>   var treemap = new google.vis**ualization.TreeMap(document.ge**
>>> tElementById('visualization'))**;
>>>   treemap.draw(data, {
>>>     minColor: 'red',
>>>     midColor: '#ddd',
>>>     maxColor: '#0d0',
>>>     maxDepth: 2,
>>>     headerHeight: 15,
>>>     fontColor: 'black',
>>>     showScale: true});
>>> }
>>>
>>> Following code raises "More than one row with the same ID (ChildA)", it
>>> should not.
>>>
>>> Olivier
>>>
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>>>
>>
>>
>>
>> --
>> Daniel LaLiberte <https://plus.google.com/100631381223468223275?prsrc=2>
>>  - 978-394-1058
>> [email protected]   562D 5CC, Cambridge MA
>> [email protected] 9 Juniper Ridge Road, Acton MA
>>
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-- 
Daniel LaLiberte <https://plus.google.com/100631381223468223275?prsrc=2>  -
978-394-1058
[email protected] <[email protected]>   562D 5CC, Cambridge MA
[email protected] <[email protected]> 9 Juniper Ridge
Road, Acton MA

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