英文:
"SyntaxError: 'super' keyword unexpected here" calling method in ES6 class
问题
class Foo {
go() {
console.log('Foo.go');
}
}
class Bar extends Foo {
go() {
console.log('Bar.go');
super(); // SyntaxError: 'super' keyword unexpected here
}
}
英文:
class Foo {
go() {
console.log('Foo.go');
}
}
class Bar extends Foo {
go() {
console.log('Bar.go');
super(); // SyntaxError: 'super' keyword unexpected here
}
}
I don't understand, this seems as simple as can be. Banged my head against this for an hour.
答案1
得分: 3
在JavaScript中，调用父类构造函数的语法是super()，但对于所有其他方法，您必须使用语法super.methodName()。
class Foo {
constructor() {
this.isFoo = true;
}
go() {
this.goFoo = true;
}
}
class Bar extends Foo {
constructor() {
super(); // 这是正确的
this.isBar = true;
}
go() {
super.go(); // 这是正确的
this.goBar = true;
}
}
const o = new Bar;
o.go();
console.log(o);
// Bar {isFoo: true, isBar: true, goFoo: true, goBar: true}
这与Java中的语法相同，但与Ruby不同，Ruby中的super()始终调用父类中同名的方法。
英文:
In JavaScript invoking the superclass constructor uses the syntax super(), but
for all other methods you must use the syntax super.methodName().
class Foo {
constructor() {
this.isFoo = true;
}
go() {
this.goFoo = true;
}
}
class Bar extends Foo {
constructor() {
super(); // This is correct
this.isBar = true;
}
go() {
super.go(); // This is correct
this.goBar = true;
}
}
const o = new Bar;
o.go();
console.log(o);
// Bar {isFoo: true, isBar: true, goFoo: true, goBar: true}
This is the same syntax as exists in Java, but different from Ruby where super() always invokes the same-named method in the parent class.
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