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Use outputs id as display_name if no display_name present, remove v3 outputs id restriction that made them have to have unique IDs from the inputs

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This commit is contained in:
Jedrzej Kosinski 2025-12-26 16:39:46 -08:00
parent 4b8b9833cb
commit 0d006f8e98
1 changed files with 3 additions and 6 deletions
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9
comfy_api/latest/_io.py
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@ -213,8 +213,9 @@ class Output(_IO_V3):
self.is_output_list = is_output_list
def as_dict(self):
display_name = self.display_name if self.display_name else self.id
return prune_dict({
"display_name": self.display_name,
"display_name": display_name,
"tooltip": self.tooltip,
"is_output_list": self.is_output_list,
})
@ -1287,16 +1288,12 @@ class Schema:
output_ids = [o.id for o in self.outputs]
input_set = set(input_ids)
output_set = set(output_ids)
issues = []
issues: list[str] = []
# verify ids are unique per list
if len(input_set) != len(input_ids):
issues.append(f"Input ids must be unique, but {[item for item, count in Counter(input_ids).items() if count > 1]} are not.")
if len(output_set) != len(output_ids):
issues.append(f"Output ids must be unique, but {[item for item, count in Counter(output_ids).items() if count > 1]} are not.")
# verify ids are unique between lists
intersection = input_set & output_set
if len(intersection) > 0:
issues.append(f"Ids must be unique between inputs and outputs, but {intersection} are not.")
if len(issues) > 0:
raise ValueError("\n".join(issues))
# validate inputs and outputs