[checklists] Ensure that issue severity is not duplicated in prenotification

Former code would produce announce text like this:

They will fix 5 security defects with severities: low, low, low, low, and low.

Author: nessita

checklists/templatetags/checklist_extras.py

@@ -53,7 +53,8 @@ def enumerate_items(items, item_formatter=None):
 
 @register.filter
 def enumerate_cves(cves, field="cve_year_number"):
-    return enumerate_items([getattr(cve, field) for cve in cves])
+    # Use `dict.fromkeys()` to preserve order while removing duplicates.
+    return enumerate_items(dict.fromkeys(getattr(cve, field) for cve in cves))
 
 
 @register.filter

checklists/tests/test_templatetags.py

@@ -78,6 +78,22 @@ def test_enumerate_cves_with_severity(self):
         result = enumerate_cves([cve1, cve2, cve3], "severity")
         self.assertEqual(result, "high, medium, and low")
 
+    def test_enumerate_cves_deduplicates_severity(self):
+        cves = [
+            self.factory.make_security_issue(severity="low", releases=[])
+            for _ in range(5)
+        ]
+        result = enumerate_cves(cves, "severity")
+        self.assertEqual(result, "low")
+
+    def test_enumerate_cves_deduplicates_mixed_severity(self):
+        cves = [
+            self.factory.make_security_issue(severity=s, releases=[])
+            for s in ("low", "moderate", "low", "moderate", "high")
+        ]
+        result = enumerate_cves(cves, "severity")
+        self.assertEqual(result, "low, moderate, and high")
+
     def test_enumerate_cves_single(self):
         cve = self.factory.make_security_issue(cve_year_number="CVE-2024-12345")
         result = enumerate_cves([cve])