SyntaxError: use of super property/member accesses only valid within methods or eval code within methods

The JavaScript exception "use of super property/member accesses only valid within methods or eval code within methods" occurs when the {{jsxref("Operators/super", "super.x")}} or super[x] syntax is used outside of a method.

Message#

SyntaxError: 'super' keyword unexpected here (V8-based)
SyntaxError: use of super property accesses only valid within methods or eval code within methods (Firefox)
SyntaxError: super is not valid in this context. (Safari)

Error type#

{{jsxref("SyntaxError")}}

What went wrong?#

The super.x syntax is used to access properties on the prototype of the current object. It can be used in methods of both object literals and classes, field initializers, and static initialization blocks, but not in other contexts.

Examples#

Invalid cases#

You can't use super.x outside of a method in an object:

```js example-bad const obj = { proto: { x: 1 }, x: super.x, // SyntaxError: use of super property accesses only valid within methods or eval code within methods };

You can't use `super.x` in a function, even if that function has the effect of being a method:

```js example-bad
function getX() {
  return super.x; // SyntaxError: use of super property accesses only valid within methods or eval code within methods
}

const obj = {
  getX,
  getX2: function () {
    return super.x; // SyntaxError: use of super property accesses only valid within methods or eval code within methods
  },
};

class Derived extends Base {
  getX = () => super.x;
}

Valid cases#

You can use super.x in a method:

```js example-good class Base { x = 1; }

class Derived extends Base { getX() { return super.x; } }

You can use `super.x` in a field initializer:

```js example-good
class Derived extends Base {
  x = super.x;
}

You can use super.x in object methods too:

js example-good const obj = { __proto__: { x: 1 }, getX() { return super.x; }, };

See also#

• Classes
• {{jsxref("Operators/super", "super")}}