Hi, I am going to describe the problem with an example:
I have three projects PA, PB and PC, and PA-100, PB-90 and PC-120 are Issues of these projects that are related as :
PA-100
Link to PB-90
Link to PC-120
With this scenario I have defined a structure that lists the occurrences of the PB project, and I have added a calculated column with the parent key following the "Link To" relationship, ie:
PB-Key PA-Key ...
--------------------------- ...
PB-90 PA-100 ...
...
What I want now is to add another calculated column to list the sibling Issue (the Issue with the same parent), to have a list like:
PB-Key PA-Key PC-Key ...
--------------------------------- ...
PB-90 PA-100 PC-120 ...
...
I have tried different schemes but in the end I don't know how to do it, I even wonder if it is possible.
Could someone help me?
Hi Victor,
The short answer is yes this is possible, but it did end up being rather complicated to implement. Here is the code I came up with:
if !count_sub_items: // if we are at a leaf (no children)
with child_keys = PARENT{ array#subtree{key}.FILTER($ != this.key) }: // grab the child keys from the parents perspective
child_keys.FILTER($ != this.key) // remove this issues key from the list (leaving siblings)
If you copy and paste this into a formula it won't recognize the first variable count_sub_items, however this is an attribute you can find from the list when you click it. The other variables should be picked up directly.
Cheers,
Nick [ALM Works]
Thanks Nick, but it doesn't work, at least, let's say the output is not what I expected.
What I have with this formula, following with the example, are the keys of all the PB elements present in the structure, i.e.
PB-Key PA-Key PC-Key ...
-------------------------------------- ...
PB-90 PA-100 PB-92, PB-98 ...
PB-92 PA-110 PB-90, PB-98 ...
PB-98 PA-130 PB-90, PB-92 ...
And this is not the idea. What I need in the PC-Key column, is the Key inside the PC project.
I have tried to get each parent Issue with this:
issuelinks.filter($.source.project="PA").source
And with this, I can get the second column (PA-Key), but I have not been able to access the children of the PA project to get the keys inside the PC project, that is what I need.
Cheers,
Víctor
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Now, I'm trying this:
With p=issuelinks.filter($.source.project="PA").source:
With q=issuelinks.filter(p.destination.project="PC").destination:
q.key
But it doesn't work, the answer is: "VALUE!"
What is the problem here?, What is wrong in this expression?
Regards,
Víctor
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Hello @Victor Crespo ,
The problem with the expression was the destination will not recognize the variable p in that way. I'm not 100% sure I am clear on the goal, but perhaps this adjustment will return what you are hoping for:
with p=issuelinks.filter($.source.project="PA").source:
with q=p.filter($.destination.project="PC").destination:
q.key
Please let me know if it helps. Please also feel free to contact our support team directly via email support@almworks.com or customer portal support.almworks.com.
Best,
David
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