From e700d72550a495f7b85a67350bd5f1df3eea960a Mon Sep 17 00:00:00 2001
From: GuinsooRocky <guinsoo@163.com>
Date: Fri, 8 May 2026 20:38:51 +0900
Subject: [PATCH 1/2] fix: getNodeRef should not fallback to element.ref in
 React 19

Regression of #593: propertyIsEnumerable('ref') returns false when the
element has no ref prop, causing fallback to element.ref which triggers
React 19's deprecation warning. Restore the version-based branch
introduced in #545.
---
 src/ref.ts            | 12 +++++++-----
 tests/ref-19.test.tsx | 10 ++++++++++
 2 files changed, 17 insertions(+), 5 deletions(-)

diff --git a/src/ref.ts b/src/ref.ts
index fd9513b9..dd289383 100644
--- a/src/ref.ts
+++ b/src/ref.ts
@@ -90,17 +90,19 @@ export const supportNodeRef = <T = any>(
 /**
  * In React 19. `ref` is not a property from node.
  * But a property from `props.ref`.
- * To check if `props.ref` exist or fallback to `ref`.
+ * In < React 19, `ref` lives on the element itself (`element.ref`).
  */
 export const getNodeRef: <T = any>(
   node: React.ReactNode,
 ) => React.Ref<T> | null = node => {
   if (node && isReactElement(node)) {
     const ele = node as any;
-
-    // Source from:
-    // https://github.com/mui/material-ui/blob/master/packages/mui-utils/src/getReactNodeRef/getReactNodeRef.ts
-    return ele.props.propertyIsEnumerable('ref') ? ele.props.ref : ele.ref;
+    // React 19: `ref` is a regular prop and reading `element.ref` triggers a
+    // deprecation warning, even when the prop is absent. Branch on the React
+    // major version to avoid touching `element.ref` on React 19+.
+    return ReactMajorVersion >= 19
+      ? (ele.props.ref ?? null)
+      : ele.ref;
   }
   return null;
 };
diff --git a/tests/ref-19.test.tsx b/tests/ref-19.test.tsx
index b51d2ce4..906577f6 100644
--- a/tests/ref-19.test.tsx
+++ b/tests/ref-19.test.tsx
@@ -44,6 +44,16 @@ describe('ref: React 19', () => {
     expect(errSpy).not.toHaveBeenCalled();
   });
 
+  it('getNodeRef on element without ref does not access element.ref', () => {
+    // Regression: previous propertyIsEnumerable('ref') check fell back to
+    // `element.ref` for elements rendered without an explicit ref prop, which
+    // triggers React 19's "Accessing element.ref was removed" warning.
+    const node = <div className="no-ref" />;
+
+    expect(getNodeRef(node)).toBeNull();
+    expect(errSpy).not.toHaveBeenCalled();
+  });
+
   it('useComposeRef', () => {
     const Demo: React.FC<React.PropsWithChildren> = ({ children }) => {
       const ref = React.useRef<HTMLDivElement>(null);

From fabcc48bc7616f0a37a688127b371b6eab3feb6a Mon Sep 17 00:00:00 2001
From: GuinsooRocky <winebaughcazaree@gmail.com>
Date: Fri, 8 May 2026 23:06:08 +0900
Subject: [PATCH 2/2] fix: align getNodeRef legacy branch return type with `??
 null`

Address @gemini-code-assist review on #759: ensure both branches return
`React.Ref<T> | null` for consumers of this utility, even on the legacy
React (<19) path where `element.ref` could theoretically be undefined.
---
 src/ref.ts | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/src/ref.ts b/src/ref.ts
index dd289383..06fd5741 100644
--- a/src/ref.ts
+++ b/src/ref.ts
@@ -102,7 +102,7 @@ export const getNodeRef: <T = any>(
     // major version to avoid touching `element.ref` on React 19+.
     return ReactMajorVersion >= 19
       ? (ele.props.ref ?? null)
-      : ele.ref;
+      : (ele.ref ?? null);
   }
   return null;
 };