如何解决打字稿中的隐式任何类型索引类型错误？

Question

如何解决此错误

• “元素隐式具有'any'类型，因为类型'any'的表达式不能用于索引类型吗？“

此错误背后的原因是什么？什么是索引类型？代码在下面给出

const getPast = action => {
  const presentToPast = {
    approve: 'approved',
    reject: 'rejected',
    edit: 'edited',
    create: 'created'
  };

  return presentToPast[action];
};

Answer 1

欢迎使用Sayan stackoverflow

代码上的问题是您使用的是未键入的参数（操作），您可以尝试：

const getPast = ( action: string ) => {
  const presentToPast = {
    approve: 'approved',
    reject: 'rejected',
    edit: 'edited',
    create: 'created'
  };

  return presentToPast[action];
};

console.log(getPast('approve'));// gonna log : approved

Answer 2

我会为presentToPast创建自定义类型

type presentToPastType = {
  approve: 'approved',
  reject: 'rejected',
  edit: 'edited',
  create: 'created',
};

const getPast = (action: keyof presentToPastType) => {
  const presentToPast = {
    approve: 'approved',
    reject: 'rejected',
    edit: 'edited',
    create: 'created',
  };

  return presentToPast[action];
};

或解决方法将是

const getPast = (action: keyof typeof presentToPast) => {
  const presentToPast = {
    approve: 'approved',
    reject: 'rejected',
    edit: 'edited',
    create: 'created',
  };

  return presentToPast[action];
};