Number of labels & labels by numeric reference

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question

martin.j asked Aug 15, '17 Matthew Gillespie commented Dec 7, '20

Number of labels & labels by numeric reference

With the new dot syntax for labels it has become pretty easy to access a label by name, but is it still possible to access it by number? If I want to get the last label on an item, I would have to know how many labels the item has in total and be able to use that as the index.

But apparently the LabelsArray datatype does not support the Length property nor will it accept anything other than a string name. It doesn't seem to be possible to get a label by its index unless I use the deprecated getlabels functions.

So how can I get the total number of labels on an object?

And how can I access the label with its index?

labels dot syntax

Answer 1 · 2017-08-15T09:39:03Z

Joerg Vogel answered Aug 15, '17 Matthew Gillespie commented Dec 7, '20

You mean this

obj.attrs.labels.subnodes[num rank]

?

access-label-by-rank.jpg (141.4 KiB)

· 6

Patrick Cloutier commented · Dec 07, 2020 at 07:24 PM

Hello, I'm trying to do the same which is referencing labels by index but on token labels.

I've tried these but they don't work:

print(token.labels[3].value);
print(token.attrs.labels[3].value);
print(token.attrs.labels.subnodes[3].value);

What am I missing?

Thanks,

0 ·

Matthew Gillespie ♦♦ Patrick Cloutier commented · Dec 07, 2020 at 09:42 PM

@Patrick Cloutier

Accessing token labels by rank is not officially supported because not all the labels a token can read or write are stored on the token itself. Many times these token labels are stored on the token's parent or grandparent token.

3 ·

Joerg Vogel Patrick Cloutier commented · Dec 07, 2020 at 09:06 PM

@Patrick Cloutier, this question is related to deleting labels at a token, because you want to get access to the node structure. I came up with a source code to achieve this in my answer here. From this code is it not to far to get what you want. The main hook is to cast the token into a treenode to get access to the treenode structure.

0 ·

Patrick Cloutier Joerg Vogel commented · Dec 07, 2020 at 09:09 PM

Thanks Jorg.

0 ·

Joerg Vogel commented · Aug 15, 2017 at 09:44 AM

Or are looking for

obj.attrs.labels.find("/1");

?

obj.attrs.labels.last.name;

-1 ·

access-label-by-rank-and-find.jpg (144.7 KiB)

martin.j Joerg Vogel commented · Aug 15, 2017 at 11:41 AM

Thanks Jörg, its so obvious now that I see it that I feel silly for asking. Thank you for taking the time to answer.

0 ·

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