Closed BPierrick closed 2 years ago
I don't understand the fix sorry, can you explain a bit more ? maybe a small unit test would have helped me to understand
I don't understand the fix sorry, can you explain a bit more ? maybe a small unit test would have helped me to understand
The problem was not related to the text length, but to the useIsNodePending
algorithm.
pending
.value[x]
and that the attribute one is valueBoolean
, the node's path is not part of attribute's path.
value[x]
and attribute's path is valueQuantity.value
valueQuantity.value
.
value[x]
has a child path valueQuantity
that is included in attribute's path valueQuantity.value
ok thanks for the clarification. I have one question though: don't you think that multiple attributes can start with the same string without sharing the same ancestor ? Sorry I don't have a real world example but I can make up one:
Observation.value[x]
Observation.valueAbsentReason
(does not exist for real but imagine)
they start with the same path Observation.value
yet they do not share a common ancestor. wouldn't that be a problem ?This example wouldn't raise any issue in my opinion. Moreover I'm not sure I completely got it because Observation.value[x]
and Observation.valueAbsentReason
do share a common ancestor, which is Observation.value[x]
right..?
This one though would be a problem :
Observation.value
Observation.valueAbsentReason
Observation.value
is NOT an an ancestor of Observation.valueAbsentReason
.I don't really identify any other edge case. And I don't think this one is likely to happen..?
no I was talking of the case where Observation.value[x]
and Observation.valueAbsentReason
would not share a common ancestor (be 2 different attributes) --> the same case as your example.
I don't know if this is likely to happen but I would say it's possible
Well yes, in that case if Observation.valueAbsentReason
is not a choice of Observation.value[x]
, then this would lead to a bug.
I think either we consider this algorithm and we suppose this would never happen. Or we would need to perform a tree search to effectively check if the nodes are common ancestors, but this solution would be more expensive.
Fixes
Definition of Done