diff --git a/docs/docs/changelog.md b/docs/docs/changelog.md
index c97f28c..21313b9 100644
--- a/docs/docs/changelog.md
+++ b/docs/docs/changelog.md
@@ -16,6 +16,10 @@ and this project adheres to [Semantic Versioning](https://semver.org/spec/v2.0.0
 
 - You can now provide 'time' to the show field on a query. This will only render the time of the task (unless the end of the task is on a different day than the start).
 
+### 🐛 Bug Fixes
+
+- Date grouping now correctly groups tasks on the same day together regardless of whether they have times
+
 ## v2.3.0 (2025-12-06)
 
 ### ✨ Features
diff --git a/plugin/src/data/transformations/grouping.test.ts b/plugin/src/data/transformations/grouping.test.ts
index fd00c8b..0a24b1f 100644
--- a/plugin/src/data/transformations/grouping.test.ts
+++ b/plugin/src/data/transformations/grouping.test.ts
@@ -465,6 +465,30 @@ describe("group by date", () => {
         },
       ],
     },
+    {
+      description: "should group tasks on same due date but different times together",
+      input: [
+        makeTask("a", {
+          due: makeDueDate("2024-01-12"),
+        }),
+        makeTask("b", {
+          due: makeDueDate("2024-01-12T09:00:00"),
+        }),
+      ],
+      expected: [
+        {
+          header: "Jan 12 ‧ Friday",
+          tasks: [
+            makeTask("a", {
+              due: makeDueDate("2024-01-12"),
+            }),
+            makeTask("b", {
+              due: makeDueDate("2024-01-12T09:00:00"),
+            }),
+          ],
+        },
+      ],
+    },
   ];
 
   for (const tc of testcases) {
diff --git a/plugin/src/data/transformations/grouping.ts b/plugin/src/data/transformations/grouping.ts
index 3403f6f..2420177 100644
--- a/plugin/src/data/transformations/grouping.ts
+++ b/plugin/src/data/transformations/grouping.ts
@@ -154,7 +154,8 @@ function groupByDate(tasks: Task[]): GroupedTasks[] {
       return "Overdue";
     }
 
-    return task.due.date;
+    // Discard any time component for grouping purposes
+    return task.due.date.split("T")[0];
   });
   const groups = Array.from(dates.entries());
   groups.sort((a, b) => {