javaScript、react-native、super.method 未定义

Question

反应原生。在子类中， super.method 未定义。为什么父类方法在super中不可见？

export class ParentClass {
    constructor(field) {
        this.field = field;
    }

    protected ruleMatch = (type: string): string => {
        return type;
    }

    onGetRule = (type: string) => {
        return this.ruleMatch(type);
    }
}

export class ChildrenClass extends ParentClass {
    // extends method of parent Class
    onGetRule = (type: string) => {
        if(type === 'typeA') {
            return 'anything';
        }
        return super.onGetRule(type); // TypeError: Cannot read property 'call' of undefined
    }
}

Answer 1

您将方法onGetRule 定义为字段声明。将onGetRule-method 的语法改为

export class ParentClass {
    constructor(field) {
        this.field = field;
    }

    protected ruleMatch = (type: string): string => {
        return type;
    }

    onGetRule(type: string) {
        return this.ruleMatch(type);
    }
}

export class ChildrenClass extends ParentClass {
    // extends method of parent Class
    onGetRule(type: string) {
        if(type === 'typeA') {
            return 'anything';
        }
        return super.onGetRule(type); // TypeError: Cannot read property 'call' of undefined
    }
}

字段声明在每个类实例上创建道具，而不是像类方法那样使用原型链。

是这样的：

class ParentClass {
  constructor(field) {
    this.field = field;
    this.onGetRule = (type) => {
        return type;
    }
  }
}

所以super.onGetRule 将是未定义的，因为它在原型链中丢失了。