Javascript/Typescript - 从对象中删除

Question

如何删除包含 filter1 的所有对象？

const obj = {
  0: {
    'filter1': {
      'key1': {
        'email1': 'email1'
      }
    }
  },
  1: {
    'filter1': {
      'key12': {
        'email12': 'email12'
      }
    }
  },
  2: {
    'filter2': {
      'key2': {
        'email2': 'email2'
      }
    }
  }
}

Answer 1

您可以使用reduce获取一个新对象:

const newObject = Object.keys(obj).reduce((newObj, key) => {
  if (!obj[key].filter1) {
    newObj[key] = obj[key];
  }

  return newObj;
}, {});

const obj = {
  0: {
    'filter1': {
      'key1': {
        'email1': 'email1'
      }
    }
  },
  1: {
    'filter1': {
      'key12': {
        'email12': 'email12'
      }
    }
  },
  2: {
    'filter2': {
      'key2': {
        'email2': 'email2'
      }
    }
  }
};

const newObject = Object.keys(obj).reduce((newObj, key) => {
  if (!obj[key].filter1) {
    newObj[key] = obj[key];
  }

  return newObj;
}, {});

console.log(newObject);

或者如 @zerkms说，使用 entries :

const entries = Object.entries(obj).reduce((newObj, [key, val]) => {
  if (val.filter1) {
    return newObj;
  }

  return {
    ...newObj,
    [key]: val,
  }
}, {});

const obj = {
  0: {
    'filter1': {
      'key1': {
        'email1': 'email1'
      }
    }
  },
  1: {
    'filter1': {
      'key12': {
        'email12': 'email12'
      }
    }
  },
  2: {
    'filter2': {
      'key2': {
        'email2': 'email2'
      }
    }
  }
};

const newObject = Object.entries(obj).reduce((newObj, [key, val]) => {
  if (val.filter1) {
    return newObj;
  }

  return {
    ...newObj,
    [key]: val,
  }
}, {});

console.log(newObject);