Match the correct clause when interface position is zero
Change-Id: I6eacc818c1596442295d39ff704fe56dea319c11
Signed-off-by: aticig <gulsum.atici@canonical.com>
diff --git a/NG-RO/osm_ng_ro/ns.py b/NG-RO/osm_ng_ro/ns.py
index 096d4df..a7c1562 100644
--- a/NG-RO/osm_ng_ro/ns.py
+++ b/NG-RO/osm_ng_ro/ns.py
@@ -1087,7 +1087,11 @@
# If the position info is provided for all the interfaces, it will be sorted
# according to position number ascendingly.
- if all(i.get("position") for i in target_vdu["interfaces"]):
+ if all(
+ i.get("position") + 1
+ for i in target_vdu["interfaces"]
+ if i.get("position") is not None
+ ):
sorted_interfaces = sorted(
target_vdu["interfaces"],
key=lambda x: (x.get("position") is None, x.get("position")),
@@ -1097,7 +1101,11 @@
# If the position info is provided for some interfaces but not all of them, the interfaces
# which has specific position numbers will be placed and others' positions will not be taken care.
else:
- if any(i.get("position") for i in target_vdu["interfaces"]):
+ if any(
+ i.get("position") + 1
+ for i in target_vdu["interfaces"]
+ if i.get("position") is not None
+ ):
n = len(target_vdu["interfaces"])
sorted_interfaces = [-1] * n
k, m = 0, 0